P. 282 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 28101-28200   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	mulsproplem4 28101*	Lemma for surreal multiplication. Under the inductive hypothesis, the product of a member of the old set of 𝐴 and a member of the old set of 𝐵 is a surreal number. (Contributed by Scott Fenton, 4-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝑋 ∈ ( O ‘( bday ‘𝐴)))    &   ⊢ (𝜑 → 𝑌 ∈ ( O ‘( bday ‘𝐵)))    ⇒   ⊢ (𝜑 → (𝑋 ·s 𝑌) ∈ No )
Theorem	mulsproplem5 28102*	Lemma for surreal multiplication. Show one of the inequalities involved in surreal multiplication's cuts. (Contributed by Scott Fenton, 4-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝑃 ∈ ( L ‘𝐴))    &   ⊢ (𝜑 → 𝑄 ∈ ( L ‘𝐵))    &   ⊢ (𝜑 → 𝑇 ∈ ( L ‘𝐴))    &   ⊢ (𝜑 → 𝑈 ∈ ( R ‘𝐵))    ⇒   ⊢ (𝜑 → (((𝑃 ·s 𝐵) +s (𝐴 ·s 𝑄)) -s (𝑃 ·s 𝑄)) <s (((𝑇 ·s 𝐵) +s (𝐴 ·s 𝑈)) -s (𝑇 ·s 𝑈)))
Theorem	mulsproplem6 28103*	Lemma for surreal multiplication. Show one of the inequalities involved in surreal multiplication's cuts. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝑃 ∈ ( L ‘𝐴))    &   ⊢ (𝜑 → 𝑄 ∈ ( L ‘𝐵))    &   ⊢ (𝜑 → 𝑉 ∈ ( R ‘𝐴))    &   ⊢ (𝜑 → 𝑊 ∈ ( L ‘𝐵))    ⇒   ⊢ (𝜑 → (((𝑃 ·s 𝐵) +s (𝐴 ·s 𝑄)) -s (𝑃 ·s 𝑄)) <s (((𝑉 ·s 𝐵) +s (𝐴 ·s 𝑊)) -s (𝑉 ·s 𝑊)))
Theorem	mulsproplem7 28104*	Lemma for surreal multiplication. Show one of the inequalities involved in surreal multiplication's cuts. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝑅 ∈ ( R ‘𝐴))    &   ⊢ (𝜑 → 𝑆 ∈ ( R ‘𝐵))    &   ⊢ (𝜑 → 𝑇 ∈ ( L ‘𝐴))    &   ⊢ (𝜑 → 𝑈 ∈ ( R ‘𝐵))    ⇒   ⊢ (𝜑 → (((𝑅 ·s 𝐵) +s (𝐴 ·s 𝑆)) -s (𝑅 ·s 𝑆)) <s (((𝑇 ·s 𝐵) +s (𝐴 ·s 𝑈)) -s (𝑇 ·s 𝑈)))
Theorem	mulsproplem8 28105*	Lemma for surreal multiplication. Show one of the inequalities involved in surreal multiplication's cuts. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝑅 ∈ ( R ‘𝐴))    &   ⊢ (𝜑 → 𝑆 ∈ ( R ‘𝐵))    &   ⊢ (𝜑 → 𝑉 ∈ ( R ‘𝐴))    &   ⊢ (𝜑 → 𝑊 ∈ ( L ‘𝐵))    ⇒   ⊢ (𝜑 → (((𝑅 ·s 𝐵) +s (𝐴 ·s 𝑆)) -s (𝑅 ·s 𝑆)) <s (((𝑉 ·s 𝐵) +s (𝐴 ·s 𝑊)) -s (𝑉 ·s 𝑊)))
Theorem	mulsproplem9 28106*	Lemma for surreal multiplication. Show that the cut involved in surreal multiplication makes sense. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → ({𝑔 ∣ ∃𝑝 ∈ ( L ‘𝐴)∃𝑞 ∈ ( L ‘𝐵)𝑔 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))} ∪ {ℎ ∣ ∃𝑟 ∈ ( R ‘𝐴)∃𝑠 ∈ ( R ‘𝐵)ℎ = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))}) <<s ({𝑖 ∣ ∃𝑡 ∈ ( L ‘𝐴)∃𝑢 ∈ ( R ‘𝐵)𝑖 = (((𝑡 ·s 𝐵) +s (𝐴 ·s 𝑢)) -s (𝑡 ·s 𝑢))} ∪ {𝑗 ∣ ∃𝑣 ∈ ( R ‘𝐴)∃𝑤 ∈ ( L ‘𝐵)𝑗 = (((𝑣 ·s 𝐵) +s (𝐴 ·s 𝑤)) -s (𝑣 ·s 𝑤))}))
Theorem	mulsproplem10 28107*	Lemma for surreal multiplication. State the cut properties of surreal multiplication. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) ∈ No ∧ ({𝑔 ∣ ∃𝑝 ∈ ( L ‘𝐴)∃𝑞 ∈ ( L ‘𝐵)𝑔 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))} ∪ {ℎ ∣ ∃𝑟 ∈ ( R ‘𝐴)∃𝑠 ∈ ( R ‘𝐵)ℎ = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))}) <<s {(𝐴 ·s 𝐵)} ∧ {(𝐴 ·s 𝐵)} <<s ({𝑖 ∣ ∃𝑡 ∈ ( L ‘𝐴)∃𝑢 ∈ ( R ‘𝐵)𝑖 = (((𝑡 ·s 𝐵) +s (𝐴 ·s 𝑢)) -s (𝑡 ·s 𝑢))} ∪ {𝑗 ∣ ∃𝑣 ∈ ( R ‘𝐴)∃𝑤 ∈ ( L ‘𝐵)𝑗 = (((𝑣 ·s 𝐵) +s (𝐴 ·s 𝑤)) -s (𝑣 ·s 𝑤))})))
Theorem	mulsproplem11 28108*	Lemma for surreal multiplication. Under the inductive hypothesis, demonstrate closure of surreal multiplication. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → (𝐴 ·s 𝐵) ∈ No )
Theorem	mulsproplem12 28109*	Lemma for surreal multiplication. Demonstrate the second half of the inductive statement assuming 𝐶 and 𝐷 are not the same age and 𝐸 and 𝐹 are not the same age. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐷 ∈ No )    &   ⊢ (𝜑 → 𝐸 ∈ No )    &   ⊢ (𝜑 → 𝐹 ∈ No )    &   ⊢ (𝜑 → 𝐶 <s 𝐷)    &   ⊢ (𝜑 → 𝐸 <s 𝐹)    &   ⊢ (𝜑 → (( bday ‘𝐶) ∈ ( bday ‘𝐷) ∨ ( bday ‘𝐷) ∈ ( bday ‘𝐶)))    &   ⊢ (𝜑 → (( bday ‘𝐸) ∈ ( bday ‘𝐹) ∨ ( bday ‘𝐹) ∈ ( bday ‘𝐸)))    ⇒   ⊢ (𝜑 → ((𝐶 ·s 𝐹) -s (𝐶 ·s 𝐸)) <s ((𝐷 ·s 𝐹) -s (𝐷 ·s 𝐸)))
Theorem	mulsproplem13 28110*	Lemma for surreal multiplication. Remove the restriction on 𝐶 and 𝐷 from mulsproplem12 28109. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐷 ∈ No )    &   ⊢ (𝜑 → 𝐸 ∈ No )    &   ⊢ (𝜑 → 𝐹 ∈ No )    &   ⊢ (𝜑 → 𝐶 <s 𝐷)    &   ⊢ (𝜑 → 𝐸 <s 𝐹)    &   ⊢ (𝜑 → (( bday ‘𝐸) ∈ ( bday ‘𝐹) ∨ ( bday ‘𝐹) ∈ ( bday ‘𝐸)))    ⇒   ⊢ (𝜑 → ((𝐶 ·s 𝐹) -s (𝐶 ·s 𝐸)) <s ((𝐷 ·s 𝐹) -s (𝐷 ·s 𝐸)))
Theorem	mulsproplem14 28111*	Lemma for surreal multiplication. Finally, we remove the restriction on 𝐸 and 𝐹 from mulsproplem12 28109 and mulsproplem13 28110. This completes the induction on surreal multiplication. mulsprop 28112 brings all this together technically. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (𝜑 → ∀𝑎 ∈ No ∀𝑏 ∈ No ∀𝑐 ∈ No ∀𝑑 ∈ No ∀𝑒 ∈ No ∀𝑓 ∈ No (((( bday ‘𝑎) +no ( bday ‘𝑏)) ∪ (((( bday ‘𝑐) +no ( bday ‘𝑒)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑓))) ∪ ((( bday ‘𝑐) +no ( bday ‘𝑓)) ∪ (( bday ‘𝑑) +no ( bday ‘𝑒))))) ∈ ((( bday ‘𝐴) +no ( bday ‘𝐵)) ∪ (((( bday ‘𝐶) +no ( bday ‘𝐸)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐹))) ∪ ((( bday ‘𝐶) +no ( bday ‘𝐹)) ∪ (( bday ‘𝐷) +no ( bday ‘𝐸))))) → ((𝑎 ·s 𝑏) ∈ No ∧ ((𝑐 <s 𝑑 ∧ 𝑒 <s 𝑓) → ((𝑐 ·s 𝑓) -s (𝑐 ·s 𝑒)) <s ((𝑑 ·s 𝑓) -s (𝑑 ·s 𝑒))))))    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐷 ∈ No )    &   ⊢ (𝜑 → 𝐸 ∈ No )    &   ⊢ (𝜑 → 𝐹 ∈ No )    &   ⊢ (𝜑 → 𝐶 <s 𝐷)    &   ⊢ (𝜑 → 𝐸 <s 𝐹)    ⇒   ⊢ (𝜑 → ((𝐶 ·s 𝐹) -s (𝐶 ·s 𝐸)) <s ((𝐷 ·s 𝐹) -s (𝐷 ·s 𝐸)))
Theorem	mulsprop 28112	Surreals are closed under multiplication and obey a particular ordering law. Theorem 3.4 of [Gonshor] p. 17. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ) ∧ (𝐶 ∈ No ∧ 𝐷 ∈ No ) ∧ (𝐸 ∈ No ∧ 𝐹 ∈ No )) → ((𝐴 ·s 𝐵) ∈ No ∧ ((𝐶 <s 𝐷 ∧ 𝐸 <s 𝐹) → ((𝐶 ·s 𝐹) -s (𝐶 ·s 𝐸)) <s ((𝐷 ·s 𝐹) -s (𝐷 ·s 𝐸)))))
Theorem	mulscutlem 28113*	Lemma for mulscut 28114. State the theorem with extra DV conditions. (Contributed by Scott Fenton, 7-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) ∈ No ∧ ({𝑎 ∣ ∃𝑝 ∈ ( L ‘𝐴)∃𝑞 ∈ ( L ‘𝐵)𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))} ∪ {𝑏 ∣ ∃𝑟 ∈ ( R ‘𝐴)∃𝑠 ∈ ( R ‘𝐵)𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))}) <<s {(𝐴 ·s 𝐵)} ∧ {(𝐴 ·s 𝐵)} <<s ({𝑐 ∣ ∃𝑡 ∈ ( L ‘𝐴)∃𝑢 ∈ ( R ‘𝐵)𝑐 = (((𝑡 ·s 𝐵) +s (𝐴 ·s 𝑢)) -s (𝑡 ·s 𝑢))} ∪ {𝑑 ∣ ∃𝑣 ∈ ( R ‘𝐴)∃𝑤 ∈ ( L ‘𝐵)𝑑 = (((𝑣 ·s 𝐵) +s (𝐴 ·s 𝑤)) -s (𝑣 ·s 𝑤))})))
Theorem	mulscut 28114*	Show the cut properties of surreal multiplication. (Contributed by Scott Fenton, 8-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) ∈ No ∧ ({𝑎 ∣ ∃𝑝 ∈ ( L ‘𝐴)∃𝑞 ∈ ( L ‘𝐵)𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))} ∪ {𝑏 ∣ ∃𝑟 ∈ ( R ‘𝐴)∃𝑠 ∈ ( R ‘𝐵)𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))}) <<s {(𝐴 ·s 𝐵)} ∧ {(𝐴 ·s 𝐵)} <<s ({𝑐 ∣ ∃𝑡 ∈ ( L ‘𝐴)∃𝑢 ∈ ( R ‘𝐵)𝑐 = (((𝑡 ·s 𝐵) +s (𝐴 ·s 𝑢)) -s (𝑡 ·s 𝑢))} ∪ {𝑑 ∣ ∃𝑣 ∈ ( R ‘𝐴)∃𝑤 ∈ ( L ‘𝐵)𝑑 = (((𝑣 ·s 𝐵) +s (𝐴 ·s 𝑤)) -s (𝑣 ·s 𝑤))})))
Theorem	mulscut2 28115*	Show that the cut involved in surreal multiplication is actually a cut. (Contributed by Scott Fenton, 7-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → ({𝑎 ∣ ∃𝑝 ∈ ( L ‘𝐴)∃𝑞 ∈ ( L ‘𝐵)𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))} ∪ {𝑏 ∣ ∃𝑟 ∈ ( R ‘𝐴)∃𝑠 ∈ ( R ‘𝐵)𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))}) <<s ({𝑐 ∣ ∃𝑡 ∈ ( L ‘𝐴)∃𝑢 ∈ ( R ‘𝐵)𝑐 = (((𝑡 ·s 𝐵) +s (𝐴 ·s 𝑢)) -s (𝑡 ·s 𝑢))} ∪ {𝑑 ∣ ∃𝑣 ∈ ( R ‘𝐴)∃𝑤 ∈ ( L ‘𝐵)𝑑 = (((𝑣 ·s 𝐵) +s (𝐴 ·s 𝑤)) -s (𝑣 ·s 𝑤))}))
Theorem	mulscl 28116	The surreals are closed under multiplication. Theorem 8(i) of [Conway] p. 19. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ) → (𝐴 ·s 𝐵) ∈ No )
Theorem	mulscld 28117	The surreals are closed under multiplication. Theorem 8(i) of [Conway] p. 19. (Contributed by Scott Fenton, 6-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → (𝐴 ·s 𝐵) ∈ No )
Theorem	sltmul 28118	An ordering relationship for surreal multiplication. Compare theorem 8(iii) of [Conway] p. 19. (Contributed by Scott Fenton, 5-Mar-2025.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ) ∧ (𝐶 ∈ No ∧ 𝐷 ∈ No )) → ((𝐴 <s 𝐵 ∧ 𝐶 <s 𝐷) → ((𝐴 ·s 𝐷) -s (𝐴 ·s 𝐶)) <s ((𝐵 ·s 𝐷) -s (𝐵 ·s 𝐶))))
Theorem	sltmuld 28119	An ordering relationship for surreal multiplication. Compare theorem 8(iii) of [Conway] p. 19. (Contributed by Scott Fenton, 6-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐷 ∈ No )    &   ⊢ (𝜑 → 𝐴 <s 𝐵)    &   ⊢ (𝜑 → 𝐶 <s 𝐷)    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐷) -s (𝐴 ·s 𝐶)) <s ((𝐵 ·s 𝐷) -s (𝐵 ·s 𝐶)))
Theorem	slemuld 28120	An ordering relationship for surreal multiplication. Compare theorem 8(iii) of [Conway] p. 19. (Contributed by Scott Fenton, 7-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐷 ∈ No )    &   ⊢ (𝜑 → 𝐴 ≤s 𝐵)    &   ⊢ (𝜑 → 𝐶 ≤s 𝐷)    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐷) -s (𝐴 ·s 𝐶)) ≤s ((𝐵 ·s 𝐷) -s (𝐵 ·s 𝐶)))
Theorem	mulscom 28121	Surreal multiplication commutes. Part of theorem 7 of [Conway] p. 19. (Contributed by Scott Fenton, 6-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ) → (𝐴 ·s 𝐵) = (𝐵 ·s 𝐴))
Theorem	mulscomd 28122	Surreal multiplication commutes. Part of theorem 7 of [Conway] p. 19. (Contributed by Scott Fenton, 6-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → (𝐴 ·s 𝐵) = (𝐵 ·s 𝐴))
Theorem	muls02 28123	Surreal multiplication by zero. (Contributed by Scott Fenton, 4-Feb-2025.)
⊢ (𝐴 ∈ No → ( 0s ·s 𝐴) = 0s )
Theorem	mulslid 28124	Surreal one is a left identity element for multiplication. (Contributed by Scott Fenton, 4-Feb-2025.)
⊢ (𝐴 ∈ No → ( 1s ·s 𝐴) = 𝐴)
Theorem	mulslidd 28125	Surreal one is a left identity element for multiplication. (Contributed by Scott Fenton, 14-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    ⇒   ⊢ (𝜑 → ( 1s ·s 𝐴) = 𝐴)
Theorem	mulsgt0 28126	The product of two positive surreals is positive. Theorem 9 of [Conway] p. 20. (Contributed by Scott Fenton, 6-Mar-2025.)
⊢ (((𝐴 ∈ No ∧ 0s <s 𝐴) ∧ (𝐵 ∈ No ∧ 0s <s 𝐵)) → 0s <s (𝐴 ·s 𝐵))
Theorem	mulsgt0d 28127	The product of two positive surreals is positive. Theorem 9 of [Conway] p. 20. (Contributed by Scott Fenton, 6-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐴)    &   ⊢ (𝜑 → 0s <s 𝐵)    ⇒   ⊢ (𝜑 → 0s <s (𝐴 ·s 𝐵))
Theorem	mulsge0d 28128	The product of two non-negative surreals is non-negative. (Contributed by Scott Fenton, 6-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 0s ≤s 𝐴)    &   ⊢ (𝜑 → 0s ≤s 𝐵)    ⇒   ⊢ (𝜑 → 0s ≤s (𝐴 ·s 𝐵))
Theorem	ssltmul1 28129*	One surreal set less-than relationship for cuts of 𝐴 and 𝐵. (Contributed by Scott Fenton, 7-Mar-2025.)
⊢ (𝜑 → 𝐿 <<s 𝑅)    &   ⊢ (𝜑 → 𝑀 <<s 𝑆)    &   ⊢ (𝜑 → 𝐴 = (𝐿 |s 𝑅))    &   ⊢ (𝜑 → 𝐵 = (𝑀 |s 𝑆))    ⇒   ⊢ (𝜑 → ({𝑎 ∣ ∃𝑝 ∈ 𝐿 ∃𝑞 ∈ 𝑀 𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))} ∪ {𝑏 ∣ ∃𝑟 ∈ 𝑅 ∃𝑠 ∈ 𝑆 𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))}) <<s {(𝐴 ·s 𝐵)})
Theorem	ssltmul2 28130*	One surreal set less-than relationship for cuts of 𝐴 and 𝐵. (Contributed by Scott Fenton, 7-Mar-2025.)
⊢ (𝜑 → 𝐿 <<s 𝑅)    &   ⊢ (𝜑 → 𝑀 <<s 𝑆)    &   ⊢ (𝜑 → 𝐴 = (𝐿 |s 𝑅))    &   ⊢ (𝜑 → 𝐵 = (𝑀 |s 𝑆))    ⇒   ⊢ (𝜑 → {(𝐴 ·s 𝐵)} <<s ({𝑐 ∣ ∃𝑡 ∈ 𝐿 ∃𝑢 ∈ 𝑆 𝑐 = (((𝑡 ·s 𝐵) +s (𝐴 ·s 𝑢)) -s (𝑡 ·s 𝑢))} ∪ {𝑑 ∣ ∃𝑣 ∈ 𝑅 ∃𝑤 ∈ 𝑀 𝑑 = (((𝑣 ·s 𝐵) +s (𝐴 ·s 𝑤)) -s (𝑣 ·s 𝑤))}))
Theorem	mulsuniflem 28131*	Lemma for mulsunif 28132. State the theorem with some extra distinct variable conditions. (Contributed by Scott Fenton, 8-Mar-2025.)
⊢ (𝜑 → 𝐿 <<s 𝑅)    &   ⊢ (𝜑 → 𝑀 <<s 𝑆)    &   ⊢ (𝜑 → 𝐴 = (𝐿 |s 𝑅))    &   ⊢ (𝜑 → 𝐵 = (𝑀 |s 𝑆))    ⇒   ⊢ (𝜑 → (𝐴 ·s 𝐵) = (({𝑎 ∣ ∃𝑝 ∈ 𝐿 ∃𝑞 ∈ 𝑀 𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))} ∪ {𝑏 ∣ ∃𝑟 ∈ 𝑅 ∃𝑠 ∈ 𝑆 𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))}) |s ({𝑐 ∣ ∃𝑡 ∈ 𝐿 ∃𝑢 ∈ 𝑆 𝑐 = (((𝑡 ·s 𝐵) +s (𝐴 ·s 𝑢)) -s (𝑡 ·s 𝑢))} ∪ {𝑑 ∣ ∃𝑣 ∈ 𝑅 ∃𝑤 ∈ 𝑀 𝑑 = (((𝑣 ·s 𝐵) +s (𝐴 ·s 𝑤)) -s (𝑣 ·s 𝑤))})))
Theorem	mulsunif 28132*	Surreal multiplication has the uniformity property. That is, any cuts that define 𝐴 and 𝐵 can be used in the definition of (𝐴 ·s 𝐵). Theorem 3.5 of [Gonshor] p. 18. (Contributed by Scott Fenton, 7-Mar-2025.)
⊢ (𝜑 → 𝐿 <<s 𝑅)    &   ⊢ (𝜑 → 𝑀 <<s 𝑆)    &   ⊢ (𝜑 → 𝐴 = (𝐿 |s 𝑅))    &   ⊢ (𝜑 → 𝐵 = (𝑀 |s 𝑆))    ⇒   ⊢ (𝜑 → (𝐴 ·s 𝐵) = (({𝑎 ∣ ∃𝑝 ∈ 𝐿 ∃𝑞 ∈ 𝑀 𝑎 = (((𝑝 ·s 𝐵) +s (𝐴 ·s 𝑞)) -s (𝑝 ·s 𝑞))} ∪ {𝑏 ∣ ∃𝑟 ∈ 𝑅 ∃𝑠 ∈ 𝑆 𝑏 = (((𝑟 ·s 𝐵) +s (𝐴 ·s 𝑠)) -s (𝑟 ·s 𝑠))}) |s ({𝑐 ∣ ∃𝑡 ∈ 𝐿 ∃𝑢 ∈ 𝑆 𝑐 = (((𝑡 ·s 𝐵) +s (𝐴 ·s 𝑢)) -s (𝑡 ·s 𝑢))} ∪ {𝑑 ∣ ∃𝑣 ∈ 𝑅 ∃𝑤 ∈ 𝑀 𝑑 = (((𝑣 ·s 𝐵) +s (𝐴 ·s 𝑤)) -s (𝑣 ·s 𝑤))})))
Theorem	addsdilem1 28133*	Lemma for surreal distribution. Expand the left hand side of the main expression. (Contributed by Scott Fenton, 8-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    ⇒   ⊢ (𝜑 → (𝐴 ·s (𝐵 +s 𝐶)) = ((({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)𝑎 = (((𝑥𝐿 ·s (𝐵 +s 𝐶)) +s (𝐴 ·s (𝑦𝐿 +s 𝐶))) -s (𝑥𝐿 ·s (𝑦𝐿 +s 𝐶)))} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = (((𝑥𝐿 ·s (𝐵 +s 𝐶)) +s (𝐴 ·s (𝐵 +s 𝑧𝐿))) -s (𝑥𝐿 ·s (𝐵 +s 𝑧𝐿)))}) ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)𝑎 = (((𝑥𝑅 ·s (𝐵 +s 𝐶)) +s (𝐴 ·s (𝑦𝑅 +s 𝐶))) -s (𝑥𝑅 ·s (𝑦𝑅 +s 𝐶)))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = (((𝑥𝑅 ·s (𝐵 +s 𝐶)) +s (𝐴 ·s (𝐵 +s 𝑧𝑅))) -s (𝑥𝑅 ·s (𝐵 +s 𝑧𝑅)))})) |s (({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)𝑎 = (((𝑥𝐿 ·s (𝐵 +s 𝐶)) +s (𝐴 ·s (𝑦𝑅 +s 𝐶))) -s (𝑥𝐿 ·s (𝑦𝑅 +s 𝐶)))} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = (((𝑥𝐿 ·s (𝐵 +s 𝐶)) +s (𝐴 ·s (𝐵 +s 𝑧𝑅))) -s (𝑥𝐿 ·s (𝐵 +s 𝑧𝑅)))}) ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)𝑎 = (((𝑥𝑅 ·s (𝐵 +s 𝐶)) +s (𝐴 ·s (𝑦𝐿 +s 𝐶))) -s (𝑥𝑅 ·s (𝑦𝐿 +s 𝐶)))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = (((𝑥𝑅 ·s (𝐵 +s 𝐶)) +s (𝐴 ·s (𝐵 +s 𝑧𝐿))) -s (𝑥𝑅 ·s (𝐵 +s 𝑧𝐿)))}))))
Theorem	addsdilem2 28134*	Lemma for surreal distribution. Expand the right hand side of the main expression. (Contributed by Scott Fenton, 8-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) +s (𝐴 ·s 𝐶)) = ((({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)𝑎 = ((((𝑥𝐿 ·s 𝐵) +s (𝐴 ·s 𝑦𝐿)) -s (𝑥𝐿 ·s 𝑦𝐿)) +s (𝐴 ·s 𝐶))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)𝑎 = ((((𝑥𝑅 ·s 𝐵) +s (𝐴 ·s 𝑦𝑅)) -s (𝑥𝑅 ·s 𝑦𝑅)) +s (𝐴 ·s 𝐶))}) ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = ((𝐴 ·s 𝐵) +s (((𝑥𝐿 ·s 𝐶) +s (𝐴 ·s 𝑧𝐿)) -s (𝑥𝐿 ·s 𝑧𝐿)))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = ((𝐴 ·s 𝐵) +s (((𝑥𝑅 ·s 𝐶) +s (𝐴 ·s 𝑧𝑅)) -s (𝑥𝑅 ·s 𝑧𝑅)))})) |s (({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)𝑎 = ((((𝑥𝐿 ·s 𝐵) +s (𝐴 ·s 𝑦𝑅)) -s (𝑥𝐿 ·s 𝑦𝑅)) +s (𝐴 ·s 𝐶))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)𝑎 = ((((𝑥𝑅 ·s 𝐵) +s (𝐴 ·s 𝑦𝐿)) -s (𝑥𝑅 ·s 𝑦𝐿)) +s (𝐴 ·s 𝐶))}) ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = ((𝐴 ·s 𝐵) +s (((𝑥𝐿 ·s 𝐶) +s (𝐴 ·s 𝑧𝑅)) -s (𝑥𝐿 ·s 𝑧𝑅)))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = ((𝐴 ·s 𝐵) +s (((𝑥𝑅 ·s 𝐶) +s (𝐴 ·s 𝑧𝐿)) -s (𝑥𝑅 ·s 𝑧𝐿)))}))))
Theorem	addsdilem3 28135*	Lemma for addsdi 28137. Show one of the equalities involved in the final expression. (Contributed by Scott Fenton, 9-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))(𝑥𝑂 ·s (𝐵 +s 𝐶)) = ((𝑥𝑂 ·s 𝐵) +s (𝑥𝑂 ·s 𝐶)))    &   ⊢ (𝜑 → ∀𝑦𝑂 ∈ (( L ‘𝐵) ∪ ( R ‘𝐵))(𝐴 ·s (𝑦𝑂 +s 𝐶)) = ((𝐴 ·s 𝑦𝑂) +s (𝐴 ·s 𝐶)))    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))∀𝑦𝑂 ∈ (( L ‘𝐵) ∪ ( R ‘𝐵))(𝑥𝑂 ·s (𝑦𝑂 +s 𝐶)) = ((𝑥𝑂 ·s 𝑦𝑂) +s (𝑥𝑂 ·s 𝐶)))    &   ⊢ (𝜓 → 𝑋 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴)))    &   ⊢ (𝜓 → 𝑌 ∈ (( L ‘𝐵) ∪ ( R ‘𝐵)))    ⇒   ⊢ ((𝜑 ∧ 𝜓) → (((𝑋 ·s (𝐵 +s 𝐶)) +s (𝐴 ·s (𝑌 +s 𝐶))) -s (𝑋 ·s (𝑌 +s 𝐶))) = ((((𝑋 ·s 𝐵) +s (𝐴 ·s 𝑌)) -s (𝑋 ·s 𝑌)) +s (𝐴 ·s 𝐶)))
Theorem	addsdilem4 28136*	Lemma for addsdi 28137. Show one of the equalities involved in the final expression. (Contributed by Scott Fenton, 9-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))(𝑥𝑂 ·s (𝐵 +s 𝐶)) = ((𝑥𝑂 ·s 𝐵) +s (𝑥𝑂 ·s 𝐶)))    &   ⊢ (𝜑 → ∀𝑧𝑂 ∈ (( L ‘𝐶) ∪ ( R ‘𝐶))(𝐴 ·s (𝐵 +s 𝑧𝑂)) = ((𝐴 ·s 𝐵) +s (𝐴 ·s 𝑧𝑂)))    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))∀𝑧𝑂 ∈ (( L ‘𝐶) ∪ ( R ‘𝐶))(𝑥𝑂 ·s (𝐵 +s 𝑧𝑂)) = ((𝑥𝑂 ·s 𝐵) +s (𝑥𝑂 ·s 𝑧𝑂)))    &   ⊢ (𝜓 → 𝑋 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴)))    &   ⊢ (𝜓 → 𝑍 ∈ (( L ‘𝐶) ∪ ( R ‘𝐶)))    ⇒   ⊢ ((𝜑 ∧ 𝜓) → (((𝑋 ·s (𝐵 +s 𝐶)) +s (𝐴 ·s (𝐵 +s 𝑍))) -s (𝑋 ·s (𝐵 +s 𝑍))) = ((𝐴 ·s 𝐵) +s (((𝑋 ·s 𝐶) +s (𝐴 ·s 𝑍)) -s (𝑋 ·s 𝑍))))
Theorem	addsdi 28137	Distributive law for surreal numbers. Commuted form of part of theorem 7 of [Conway] p. 19. (Contributed by Scott Fenton, 9-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐶 ∈ No ) → (𝐴 ·s (𝐵 +s 𝐶)) = ((𝐴 ·s 𝐵) +s (𝐴 ·s 𝐶)))
Theorem	addsdid 28138	Distributive law for surreal numbers. Commuted form of part of theorem 7 of [Conway] p. 19. (Contributed by Scott Fenton, 9-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    ⇒   ⊢ (𝜑 → (𝐴 ·s (𝐵 +s 𝐶)) = ((𝐴 ·s 𝐵) +s (𝐴 ·s 𝐶)))
Theorem	addsdird 28139	Distributive law for surreal numbers. Part of theorem 7 of [Conway] p. 19. (Contributed by Scott Fenton, 9-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 +s 𝐵) ·s 𝐶) = ((𝐴 ·s 𝐶) +s (𝐵 ·s 𝐶)))
Theorem	subsdid 28140	Distribution of surreal multiplication over subtraction. (Contributed by Scott Fenton, 9-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    ⇒   ⊢ (𝜑 → (𝐴 ·s (𝐵 -s 𝐶)) = ((𝐴 ·s 𝐵) -s (𝐴 ·s 𝐶)))
Theorem	subsdird 28141	Distribution of surreal multiplication over subtraction. (Contributed by Scott Fenton, 9-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 -s 𝐵) ·s 𝐶) = ((𝐴 ·s 𝐶) -s (𝐵 ·s 𝐶)))
Theorem	mulnegs1d 28142	Product with negative is negative of product. Part of theorem 7 of [Conway] p. 19. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → (( -us ‘𝐴) ·s 𝐵) = ( -us ‘(𝐴 ·s 𝐵)))
Theorem	mulnegs2d 28143	Product with negative is negative of product. Part of theorem 7 of [Conway] p. 19. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → (𝐴 ·s ( -us ‘𝐵)) = ( -us ‘(𝐴 ·s 𝐵)))
Theorem	mul2negsd 28144	Surreal product of two negatives. (Contributed by Scott Fenton, 15-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → (( -us ‘𝐴) ·s ( -us ‘𝐵)) = (𝐴 ·s 𝐵))
Theorem	mulsasslem1 28145*	Lemma for mulsass 28148. Expand the left hand side of the formula. (Contributed by Scott Fenton, 9-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) ·s 𝐶) = ((({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = ((((((𝑥𝐿 ·s 𝐵) +s (𝐴 ·s 𝑦𝐿)) -s (𝑥𝐿 ·s 𝑦𝐿)) ·s 𝐶) +s ((𝐴 ·s 𝐵) ·s 𝑧𝐿)) -s ((((𝑥𝐿 ·s 𝐵) +s (𝐴 ·s 𝑦𝐿)) -s (𝑥𝐿 ·s 𝑦𝐿)) ·s 𝑧𝐿))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = ((((((𝑥𝑅 ·s 𝐵) +s (𝐴 ·s 𝑦𝑅)) -s (𝑥𝑅 ·s 𝑦𝑅)) ·s 𝐶) +s ((𝐴 ·s 𝐵) ·s 𝑧𝐿)) -s ((((𝑥𝑅 ·s 𝐵) +s (𝐴 ·s 𝑦𝑅)) -s (𝑥𝑅 ·s 𝑦𝑅)) ·s 𝑧𝐿))}) ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = ((((((𝑥𝐿 ·s 𝐵) +s (𝐴 ·s 𝑦𝑅)) -s (𝑥𝐿 ·s 𝑦𝑅)) ·s 𝐶) +s ((𝐴 ·s 𝐵) ·s 𝑧𝑅)) -s ((((𝑥𝐿 ·s 𝐵) +s (𝐴 ·s 𝑦𝑅)) -s (𝑥𝐿 ·s 𝑦𝑅)) ·s 𝑧𝑅))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = ((((((𝑥𝑅 ·s 𝐵) +s (𝐴 ·s 𝑦𝐿)) -s (𝑥𝑅 ·s 𝑦𝐿)) ·s 𝐶) +s ((𝐴 ·s 𝐵) ·s 𝑧𝑅)) -s ((((𝑥𝑅 ·s 𝐵) +s (𝐴 ·s 𝑦𝐿)) -s (𝑥𝑅 ·s 𝑦𝐿)) ·s 𝑧𝑅))})) |s (({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = ((((((𝑥𝐿 ·s 𝐵) +s (𝐴 ·s 𝑦𝐿)) -s (𝑥𝐿 ·s 𝑦𝐿)) ·s 𝐶) +s ((𝐴 ·s 𝐵) ·s 𝑧𝑅)) -s ((((𝑥𝐿 ·s 𝐵) +s (𝐴 ·s 𝑦𝐿)) -s (𝑥𝐿 ·s 𝑦𝐿)) ·s 𝑧𝑅))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = ((((((𝑥𝑅 ·s 𝐵) +s (𝐴 ·s 𝑦𝑅)) -s (𝑥𝑅 ·s 𝑦𝑅)) ·s 𝐶) +s ((𝐴 ·s 𝐵) ·s 𝑧𝑅)) -s ((((𝑥𝑅 ·s 𝐵) +s (𝐴 ·s 𝑦𝑅)) -s (𝑥𝑅 ·s 𝑦𝑅)) ·s 𝑧𝑅))}) ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = ((((((𝑥𝐿 ·s 𝐵) +s (𝐴 ·s 𝑦𝑅)) -s (𝑥𝐿 ·s 𝑦𝑅)) ·s 𝐶) +s ((𝐴 ·s 𝐵) ·s 𝑧𝐿)) -s ((((𝑥𝐿 ·s 𝐵) +s (𝐴 ·s 𝑦𝑅)) -s (𝑥𝐿 ·s 𝑦𝑅)) ·s 𝑧𝐿))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = ((((((𝑥𝑅 ·s 𝐵) +s (𝐴 ·s 𝑦𝐿)) -s (𝑥𝑅 ·s 𝑦𝐿)) ·s 𝐶) +s ((𝐴 ·s 𝐵) ·s 𝑧𝐿)) -s ((((𝑥𝑅 ·s 𝐵) +s (𝐴 ·s 𝑦𝐿)) -s (𝑥𝑅 ·s 𝑦𝐿)) ·s 𝑧𝐿))}))))
Theorem	mulsasslem2 28146*	Lemma for mulsass 28148. Expand the right hand side of the formula. (Contributed by Scott Fenton, 9-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    ⇒   ⊢ (𝜑 → (𝐴 ·s (𝐵 ·s 𝐶)) = ((({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = (((𝑥𝐿 ·s (𝐵 ·s 𝐶)) +s (𝐴 ·s (((𝑦𝐿 ·s 𝐶) +s (𝐵 ·s 𝑧𝐿)) -s (𝑦𝐿 ·s 𝑧𝐿)))) -s (𝑥𝐿 ·s (((𝑦𝐿 ·s 𝐶) +s (𝐵 ·s 𝑧𝐿)) -s (𝑦𝐿 ·s 𝑧𝐿))))} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = (((𝑥𝐿 ·s (𝐵 ·s 𝐶)) +s (𝐴 ·s (((𝑦𝑅 ·s 𝐶) +s (𝐵 ·s 𝑧𝑅)) -s (𝑦𝑅 ·s 𝑧𝑅)))) -s (𝑥𝐿 ·s (((𝑦𝑅 ·s 𝐶) +s (𝐵 ·s 𝑧𝑅)) -s (𝑦𝑅 ·s 𝑧𝑅))))}) ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = (((𝑥𝑅 ·s (𝐵 ·s 𝐶)) +s (𝐴 ·s (((𝑦𝐿 ·s 𝐶) +s (𝐵 ·s 𝑧𝑅)) -s (𝑦𝐿 ·s 𝑧𝑅)))) -s (𝑥𝑅 ·s (((𝑦𝐿 ·s 𝐶) +s (𝐵 ·s 𝑧𝑅)) -s (𝑦𝐿 ·s 𝑧𝑅))))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = (((𝑥𝑅 ·s (𝐵 ·s 𝐶)) +s (𝐴 ·s (((𝑦𝑅 ·s 𝐶) +s (𝐵 ·s 𝑧𝐿)) -s (𝑦𝑅 ·s 𝑧𝐿)))) -s (𝑥𝑅 ·s (((𝑦𝑅 ·s 𝐶) +s (𝐵 ·s 𝑧𝐿)) -s (𝑦𝑅 ·s 𝑧𝐿))))})) |s (({𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = (((𝑥𝐿 ·s (𝐵 ·s 𝐶)) +s (𝐴 ·s (((𝑦𝐿 ·s 𝐶) +s (𝐵 ·s 𝑧𝑅)) -s (𝑦𝐿 ·s 𝑧𝑅)))) -s (𝑥𝐿 ·s (((𝑦𝐿 ·s 𝐶) +s (𝐵 ·s 𝑧𝑅)) -s (𝑦𝐿 ·s 𝑧𝑅))))} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ ( L ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = (((𝑥𝐿 ·s (𝐵 ·s 𝐶)) +s (𝐴 ·s (((𝑦𝑅 ·s 𝐶) +s (𝐵 ·s 𝑧𝐿)) -s (𝑦𝑅 ·s 𝑧𝐿)))) -s (𝑥𝐿 ·s (((𝑦𝑅 ·s 𝐶) +s (𝐵 ·s 𝑧𝐿)) -s (𝑦𝑅 ·s 𝑧𝐿))))}) ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ ( L ‘𝐵)∃𝑧𝐿 ∈ ( L ‘𝐶)𝑎 = (((𝑥𝑅 ·s (𝐵 ·s 𝐶)) +s (𝐴 ·s (((𝑦𝐿 ·s 𝐶) +s (𝐵 ·s 𝑧𝐿)) -s (𝑦𝐿 ·s 𝑧𝐿)))) -s (𝑥𝑅 ·s (((𝑦𝐿 ·s 𝐶) +s (𝐵 ·s 𝑧𝐿)) -s (𝑦𝐿 ·s 𝑧𝐿))))} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ ( R ‘𝐵)∃𝑧𝑅 ∈ ( R ‘𝐶)𝑎 = (((𝑥𝑅 ·s (𝐵 ·s 𝐶)) +s (𝐴 ·s (((𝑦𝑅 ·s 𝐶) +s (𝐵 ·s 𝑧𝑅)) -s (𝑦𝑅 ·s 𝑧𝑅)))) -s (𝑥𝑅 ·s (((𝑦𝑅 ·s 𝐶) +s (𝐵 ·s 𝑧𝑅)) -s (𝑦𝑅 ·s 𝑧𝑅))))}))))
Theorem	mulsasslem3 28147*	Lemma for mulsass 28148. Demonstrate the central equality. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ 𝑃 ⊆ (( L ‘𝐴) ∪ ( R ‘𝐴))    &   ⊢ 𝑄 ⊆ (( L ‘𝐵) ∪ ( R ‘𝐵))    &   ⊢ 𝑅 ⊆ (( L ‘𝐶) ∪ ( R ‘𝐶))    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))∀𝑦𝑂 ∈ (( L ‘𝐵) ∪ ( R ‘𝐵))∀𝑧𝑂 ∈ (( L ‘𝐶) ∪ ( R ‘𝐶))((𝑥𝑂 ·s 𝑦𝑂) ·s 𝑧𝑂) = (𝑥𝑂 ·s (𝑦𝑂 ·s 𝑧𝑂)))    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))∀𝑦𝑂 ∈ (( L ‘𝐵) ∪ ( R ‘𝐵))((𝑥𝑂 ·s 𝑦𝑂) ·s 𝐶) = (𝑥𝑂 ·s (𝑦𝑂 ·s 𝐶)))    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))∀𝑧𝑂 ∈ (( L ‘𝐶) ∪ ( R ‘𝐶))((𝑥𝑂 ·s 𝐵) ·s 𝑧𝑂) = (𝑥𝑂 ·s (𝐵 ·s 𝑧𝑂)))    &   ⊢ (𝜑 → ∀𝑦𝑂 ∈ (( L ‘𝐵) ∪ ( R ‘𝐵))∀𝑧𝑂 ∈ (( L ‘𝐶) ∪ ( R ‘𝐶))((𝐴 ·s 𝑦𝑂) ·s 𝑧𝑂) = (𝐴 ·s (𝑦𝑂 ·s 𝑧𝑂)))    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))((𝑥𝑂 ·s 𝐵) ·s 𝐶) = (𝑥𝑂 ·s (𝐵 ·s 𝐶)))    &   ⊢ (𝜑 → ∀𝑦𝑂 ∈ (( L ‘𝐵) ∪ ( R ‘𝐵))((𝐴 ·s 𝑦𝑂) ·s 𝐶) = (𝐴 ·s (𝑦𝑂 ·s 𝐶)))    &   ⊢ (𝜑 → ∀𝑧𝑂 ∈ (( L ‘𝐶) ∪ ( R ‘𝐶))((𝐴 ·s 𝐵) ·s 𝑧𝑂) = (𝐴 ·s (𝐵 ·s 𝑧𝑂)))    ⇒   ⊢ (𝜑 → (∃𝑥 ∈ 𝑃 ∃𝑦 ∈ 𝑄 ∃𝑧 ∈ 𝑅 𝑎 = ((((((𝑥 ·s 𝐵) +s (𝐴 ·s 𝑦)) -s (𝑥 ·s 𝑦)) ·s 𝐶) +s ((𝐴 ·s 𝐵) ·s 𝑧)) -s ((((𝑥 ·s 𝐵) +s (𝐴 ·s 𝑦)) -s (𝑥 ·s 𝑦)) ·s 𝑧)) ↔ ∃𝑥 ∈ 𝑃 ∃𝑦 ∈ 𝑄 ∃𝑧 ∈ 𝑅 𝑎 = (((𝑥 ·s (𝐵 ·s 𝐶)) +s (𝐴 ·s (((𝑦 ·s 𝐶) +s (𝐵 ·s 𝑧)) -s (𝑦 ·s 𝑧)))) -s (𝑥 ·s (((𝑦 ·s 𝐶) +s (𝐵 ·s 𝑧)) -s (𝑦 ·s 𝑧))))))
Theorem	mulsass 28148	Associative law for surreal multiplication. Part of theorem 7 of [Conway] p. 19. Much like the case for additive groups, this theorem together with mulscom 28121, addsdi 28137, mulsgt0 28126, and the addition theorems would make the surreals into an ordered ring except that they are a proper class. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐶 ∈ No ) → ((𝐴 ·s 𝐵) ·s 𝐶) = (𝐴 ·s (𝐵 ·s 𝐶)))
Theorem	mulsassd 28149	Associative law for surreal multiplication. Part of theorem 7 of [Conway] p. 19. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) ·s 𝐶) = (𝐴 ·s (𝐵 ·s 𝐶)))
Theorem	muls4d 28150	Rearrangement of four surreal factors. (Contributed by Scott Fenton, 16-Apr-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐷 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) ·s (𝐶 ·s 𝐷)) = ((𝐴 ·s 𝐶) ·s (𝐵 ·s 𝐷)))
Theorem	mulsunif2lem 28151*	Lemma for mulsunif2 28152. State the theorem with extra disjoint variable conditions. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐿 <<s 𝑅)    &   ⊢ (𝜑 → 𝑀 <<s 𝑆)    &   ⊢ (𝜑 → 𝐴 = (𝐿 |s 𝑅))    &   ⊢ (𝜑 → 𝐵 = (𝑀 |s 𝑆))    ⇒   ⊢ (𝜑 → (𝐴 ·s 𝐵) = (({𝑎 ∣ ∃𝑝 ∈ 𝐿 ∃𝑞 ∈ 𝑀 𝑎 = ((𝐴 ·s 𝐵) -s ((𝐴 -s 𝑝) ·s (𝐵 -s 𝑞)))} ∪ {𝑏 ∣ ∃𝑟 ∈ 𝑅 ∃𝑠 ∈ 𝑆 𝑏 = ((𝐴 ·s 𝐵) -s ((𝑟 -s 𝐴) ·s (𝑠 -s 𝐵)))}) |s ({𝑐 ∣ ∃𝑡 ∈ 𝐿 ∃𝑢 ∈ 𝑆 𝑐 = ((𝐴 ·s 𝐵) +s ((𝐴 -s 𝑡) ·s (𝑢 -s 𝐵)))} ∪ {𝑑 ∣ ∃𝑣 ∈ 𝑅 ∃𝑤 ∈ 𝑀 𝑑 = ((𝐴 ·s 𝐵) +s ((𝑣 -s 𝐴) ·s (𝐵 -s 𝑤)))})))
Theorem	mulsunif2 28152*	Alternate expression for surreal multiplication. Note from [Conway] p. 19. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐿 <<s 𝑅)    &   ⊢ (𝜑 → 𝑀 <<s 𝑆)    &   ⊢ (𝜑 → 𝐴 = (𝐿 |s 𝑅))    &   ⊢ (𝜑 → 𝐵 = (𝑀 |s 𝑆))    ⇒   ⊢ (𝜑 → (𝐴 ·s 𝐵) = (({𝑎 ∣ ∃𝑝 ∈ 𝐿 ∃𝑞 ∈ 𝑀 𝑎 = ((𝐴 ·s 𝐵) -s ((𝐴 -s 𝑝) ·s (𝐵 -s 𝑞)))} ∪ {𝑏 ∣ ∃𝑟 ∈ 𝑅 ∃𝑠 ∈ 𝑆 𝑏 = ((𝐴 ·s 𝐵) -s ((𝑟 -s 𝐴) ·s (𝑠 -s 𝐵)))}) |s ({𝑐 ∣ ∃𝑡 ∈ 𝐿 ∃𝑢 ∈ 𝑆 𝑐 = ((𝐴 ·s 𝐵) +s ((𝐴 -s 𝑡) ·s (𝑢 -s 𝐵)))} ∪ {𝑑 ∣ ∃𝑣 ∈ 𝑅 ∃𝑤 ∈ 𝑀 𝑑 = ((𝐴 ·s 𝐵) +s ((𝑣 -s 𝐴) ·s (𝐵 -s 𝑤)))})))
Theorem	sltmul2 28153	Multiplication of both sides of surreal less-than by a positive number. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (((𝐴 ∈ No ∧ 0s <s 𝐴) ∧ 𝐵 ∈ No ∧ 𝐶 ∈ No ) → (𝐵 <s 𝐶 ↔ (𝐴 ·s 𝐵) <s (𝐴 ·s 𝐶)))
Theorem	sltmul2d 28154	Multiplication of both sides of surreal less-than by a positive number. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    ⇒   ⊢ (𝜑 → (𝐴 <s 𝐵 ↔ (𝐶 ·s 𝐴) <s (𝐶 ·s 𝐵)))
Theorem	sltmul1d 28155	Multiplication of both sides of surreal less-than by a positive number. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    ⇒   ⊢ (𝜑 → (𝐴 <s 𝐵 ↔ (𝐴 ·s 𝐶) <s (𝐵 ·s 𝐶)))
Theorem	slemul2d 28156	Multiplication of both sides of surreal less-than or equal by a positive number. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    ⇒   ⊢ (𝜑 → (𝐴 ≤s 𝐵 ↔ (𝐶 ·s 𝐴) ≤s (𝐶 ·s 𝐵)))
Theorem	slemul1d 28157	Multiplication of both sides of surreal less-than or equal by a positive number. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    ⇒   ⊢ (𝜑 → (𝐴 ≤s 𝐵 ↔ (𝐴 ·s 𝐶) ≤s (𝐵 ·s 𝐶)))
Theorem	sltmulneg1d 28158	Multiplication of both sides of surreal less-than by a negative number. (Contributed by Scott Fenton, 14-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐶 <s 0s )    ⇒   ⊢ (𝜑 → (𝐴 <s 𝐵 ↔ (𝐵 ·s 𝐶) <s (𝐴 ·s 𝐶)))
Theorem	sltmulneg2d 28159	Multiplication of both sides of surreal less-than by a negative number. (Contributed by Scott Fenton, 14-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐶 <s 0s )    ⇒   ⊢ (𝜑 → (𝐴 <s 𝐵 ↔ (𝐶 ·s 𝐵) <s (𝐶 ·s 𝐴)))
Theorem	mulscan2dlem 28160	Lemma for mulscan2d 28161. Cancellation of multiplication when the right term is positive. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐶) = (𝐵 ·s 𝐶) ↔ 𝐴 = 𝐵))
Theorem	mulscan2d 28161	Cancellation of surreal multiplication when the right term is non-zero. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐶 ≠ 0s )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐶) = (𝐵 ·s 𝐶) ↔ 𝐴 = 𝐵))
Theorem	mulscan1d 28162	Cancellation of surreal multiplication when the left term is non-zero. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐶 ≠ 0s )    ⇒   ⊢ (𝜑 → ((𝐶 ·s 𝐴) = (𝐶 ·s 𝐵) ↔ 𝐴 = 𝐵))
Theorem	muls12d 28163	Commutative/associative law for surreal multiplication. (Contributed by Scott Fenton, 14-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    ⇒   ⊢ (𝜑 → (𝐴 ·s (𝐵 ·s 𝐶)) = (𝐵 ·s (𝐴 ·s 𝐶)))
Theorem	slemul1ad 28164	Multiplication of both sides of surreal less-than or equal by a non-negative number. (Contributed by Scott Fenton, 17-Apr-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s ≤s 𝐶)    &   ⊢ (𝜑 → 𝐴 ≤s 𝐵)    ⇒   ⊢ (𝜑 → (𝐴 ·s 𝐶) ≤s (𝐵 ·s 𝐶))
Theorem	sltmul12ad 28165	Comparison of the product of two positive surreals. (Contributed by Scott Fenton, 17-Apr-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐷 ∈ No )    &   ⊢ (𝜑 → 0s ≤s 𝐴)    &   ⊢ (𝜑 → 𝐴 <s 𝐵)    &   ⊢ (𝜑 → 0s ≤s 𝐶)    &   ⊢ (𝜑 → 𝐶 <s 𝐷)    ⇒   ⊢ (𝜑 → (𝐴 ·s 𝐶) <s (𝐵 ·s 𝐷))
Theorem	divsmo 28166*	Uniqueness of surreal inversion. Given a non-zero surreal 𝐴, there is at most one surreal giving a particular product. (Contributed by Scott Fenton, 10-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 𝐴 ≠ 0s ) → ∃*𝑥 ∈ No (𝐴 ·s 𝑥) = 𝐵)
Theorem	muls0ord 28167	If a surreal product is zero, one of its factors must be zero. (Contributed by Scott Fenton, 16-Apr-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) = 0s ↔ (𝐴 = 0s ∨ 𝐵 = 0s )))
Theorem	mulsne0bd 28168	The product of two non-zero surreals is non-zero. (Contributed by Scott Fenton, 16-Apr-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) ≠ 0s ↔ (𝐴 ≠ 0s ∧ 𝐵 ≠ 0s )))
15.5.4  Division
Syntax	cdivs 28169	Declare the syntax for surreal division.
class /su
Definition	df-divs 28170*	Define surreal division. This is not the definition used in the literature, but we use it here because it is technically easier to work with. (Contributed by Scott Fenton, 12-Mar-2025.)
⊢ /su = (𝑥 ∈ No , 𝑦 ∈ ( No ∖ { 0s }) ↦ (℩𝑧 ∈ No (𝑦 ·s 𝑧) = 𝑥))
Theorem	divsval 28171*	The value of surreal division. (Contributed by Scott Fenton, 12-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) → (𝐴 /su 𝐵) = (℩𝑥 ∈ No (𝐵 ·s 𝑥) = 𝐴))
Theorem	norecdiv 28172*	If a surreal has a reciprocal, then it has any division. (Contributed by Scott Fenton, 12-Mar-2025.)
⊢ (((𝐴 ∈ No ∧ 𝐴 ≠ 0s ∧ 𝐵 ∈ No ) ∧ ∃𝑥 ∈ No (𝐴 ·s 𝑥) = 1s ) → ∃𝑦 ∈ No (𝐴 ·s 𝑦) = 𝐵)
Theorem	noreceuw 28173*	If a surreal has a reciprocal, then it has unique division. (Contributed by Scott Fenton, 12-Mar-2025.)
⊢ (((𝐴 ∈ No ∧ 𝐴 ≠ 0s ∧ 𝐵 ∈ No ) ∧ ∃𝑥 ∈ No (𝐴 ·s 𝑥) = 1s ) → ∃!𝑦 ∈ No (𝐴 ·s 𝑦) = 𝐵)
Theorem	recsne0 28174*	If a surreal has a reciprocal, then it is non-zero. (Contributed by Scott Fenton, 5-Sep-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐴 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → 𝐴 ≠ 0s )
Theorem	divsmulw 28175*	Relationship between surreal division and multiplication. Weak version that does not assume reciprocals. Later, when we prove precsex 28199, we can eliminate the existence hypothesis (see divsmul 28202). (Contributed by Scott Fenton, 12-Mar-2025.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ (𝐶 ∈ No ∧ 𝐶 ≠ 0s )) ∧ ∃𝑥 ∈ No (𝐶 ·s 𝑥) = 1s ) → ((𝐴 /su 𝐶) = 𝐵 ↔ (𝐶 ·s 𝐵) = 𝐴))
Theorem	divsmulwd 28176*	Relationship between surreal division and multiplication. Weak version that does not assume reciprocals. (Contributed by Scott Fenton, 12-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐶 ≠ 0s )    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐶 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → ((𝐴 /su 𝐶) = 𝐵 ↔ (𝐶 ·s 𝐵) = 𝐴))
Theorem	divsclw 28177*	Weak division closure law. (Contributed by Scott Fenton, 12-Mar-2025.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) ∧ ∃𝑥 ∈ No (𝐵 ·s 𝑥) = 1s ) → (𝐴 /su 𝐵) ∈ No )
Theorem	divsclwd 28178*	Weak division closure law. (Contributed by Scott Fenton, 12-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐵 ≠ 0s )    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐵 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → (𝐴 /su 𝐵) ∈ No )
Theorem	divscan2wd 28179*	A weak cancellation law for surreal division. (Contributed by Scott Fenton, 13-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐵 ≠ 0s )    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐵 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → (𝐵 ·s (𝐴 /su 𝐵)) = 𝐴)
Theorem	divscan1wd 28180*	A weak cancellation law for surreal division. (Contributed by Scott Fenton, 13-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐵 ≠ 0s )    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐵 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → ((𝐴 /su 𝐵) ·s 𝐵) = 𝐴)
Theorem	sltdivmulwd 28181*	Surreal less-than relationship between division and multiplication. Weak version. (Contributed by Scott Fenton, 14-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐶 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → ((𝐴 /su 𝐶) <s 𝐵 ↔ 𝐴 <s (𝐶 ·s 𝐵)))
Theorem	sltdivmul2wd 28182*	Surreal less-than relationship between division and multiplication. Weak version. (Contributed by Scott Fenton, 14-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐶 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → ((𝐴 /su 𝐶) <s 𝐵 ↔ 𝐴 <s (𝐵 ·s 𝐶)))
Theorem	sltmuldivwd 28183*	Surreal less-than relationship between division and multiplication. Weak version. (Contributed by Scott Fenton, 14-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐶 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐶) <s 𝐵 ↔ 𝐴 <s (𝐵 /su 𝐶)))
Theorem	sltmuldiv2wd 28184*	Surreal less-than relationship between division and multiplication. Weak version. (Contributed by Scott Fenton, 14-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐶 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → ((𝐶 ·s 𝐴) <s 𝐵 ↔ 𝐴 <s (𝐵 /su 𝐶)))
Theorem	divsasswd 28185*	An associative law for surreal division. Weak version. (Contributed by Scott Fenton, 14-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐶 ≠ 0s )    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐶 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) /su 𝐶) = (𝐴 ·s (𝐵 /su 𝐶)))
Theorem	divs1 28186	A surreal divided by one is itself. (Contributed by Scott Fenton, 13-Mar-2025.)
⊢ (𝐴 ∈ No → (𝐴 /su 1s ) = 𝐴)
Theorem	precsexlemcbv 28187*	Lemma for surreal reciprocals. Change some bound variables. (Contributed by Scott Fenton, 15-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    ⇒   ⊢ 𝐹 = rec((𝑞 ∈ V ↦ ⦋(1st ‘𝑞) / 𝑚⦌⦋(2nd ‘𝑞) / 𝑠⦌⟨(𝑚 ∪ ({𝑏 ∣ ∃𝑧𝑅 ∈ ( R ‘𝐴)∃𝑤 ∈ 𝑚 𝑏 = (( 1s +s ((𝑧𝑅 -s 𝐴) ·s 𝑤)) /su 𝑧𝑅)} ∪ {𝑏 ∣ ∃𝑧𝐿 ∈ {𝑧 ∈ ( L ‘𝐴) ∣ 0s <s 𝑧}∃𝑡 ∈ 𝑠 𝑏 = (( 1s +s ((𝑧𝐿 -s 𝐴) ·s 𝑡)) /su 𝑧𝐿)})), (𝑠 ∪ ({𝑏 ∣ ∃𝑧𝐿 ∈ {𝑧 ∈ ( L ‘𝐴) ∣ 0s <s 𝑧}∃𝑤 ∈ 𝑚 𝑏 = (( 1s +s ((𝑧𝐿 -s 𝐴) ·s 𝑤)) /su 𝑧𝐿)} ∪ {𝑏 ∣ ∃𝑧𝑅 ∈ ( R ‘𝐴)∃𝑡 ∈ 𝑠 𝑏 = (( 1s +s ((𝑧𝑅 -s 𝐴) ·s 𝑡)) /su 𝑧𝑅)}))⟩), ⟨{ 0s }, ∅⟩)
Theorem	precsexlem1 28188	Lemma for surreal reciprocals. Calculate the value of the recursive left function at zero. (Contributed by Scott Fenton, 13-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    ⇒   ⊢ (𝐿‘∅) = { 0s }
Theorem	precsexlem2 28189	Lemma for surreal reciprocals. Calculate the value of the recursive right function at zero. (Contributed by Scott Fenton, 13-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    ⇒   ⊢ (𝑅‘∅) = ∅
Theorem	precsexlem3 28190*	Lemma for surreal reciprocals. Calculate the value of the recursive function at a successor. (Contributed by Scott Fenton, 12-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    ⇒   ⊢ (𝐼 ∈ ω → (𝐹‘suc 𝐼) = ⟨((𝐿‘𝐼) ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ (𝐿‘𝐼)𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ (𝑅‘𝐼)𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), ((𝑅‘𝐼) ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ (𝐿‘𝐼)𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ (𝑅‘𝐼)𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩)
Theorem	precsexlem4 28191*	Lemma for surreal reciprocals. Calculate the value of the recursive left function at a successor. (Contributed by Scott Fenton, 13-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    ⇒   ⊢ (𝐼 ∈ ω → (𝐿‘suc 𝐼) = ((𝐿‘𝐼) ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ (𝐿‘𝐼)𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ (𝑅‘𝐼)𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})))
Theorem	precsexlem5 28192*	Lemma for surreal reciprocals. Calculate the value of the recursive right function at a successor. (Contributed by Scott Fenton, 13-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    ⇒   ⊢ (𝐼 ∈ ω → (𝑅‘suc 𝐼) = ((𝑅‘𝐼) ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ (𝐿‘𝐼)𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ (𝑅‘𝐼)𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)})))
Theorem	precsexlem6 28193*	Lemma for surreal reciprocal. Show that 𝐿 is non-strictly increasing in its argument. (Contributed by Scott Fenton, 15-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    ⇒   ⊢ ((𝐼 ∈ ω ∧ 𝐽 ∈ ω ∧ 𝐼 ⊆ 𝐽) → (𝐿‘𝐼) ⊆ (𝐿‘𝐽))
Theorem	precsexlem7 28194*	Lemma for surreal reciprocal. Show that 𝑅 is non-strictly increasing in its argument. (Contributed by Scott Fenton, 15-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    ⇒   ⊢ ((𝐼 ∈ ω ∧ 𝐽 ∈ ω ∧ 𝐼 ⊆ 𝐽) → (𝑅‘𝐼) ⊆ (𝑅‘𝐽))
Theorem	precsexlem8 28195*	Lemma for surreal reciprocal. Show that the left and right functions give sets of surreals. (Contributed by Scott Fenton, 13-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐴)    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))( 0s <s 𝑥𝑂 → ∃𝑦 ∈ No (𝑥𝑂 ·s 𝑦) = 1s ))    ⇒   ⊢ ((𝜑 ∧ 𝐼 ∈ ω) → ((𝐿‘𝐼) ⊆ No ∧ (𝑅‘𝐼) ⊆ No ))
Theorem	precsexlem9 28196*	Lemma for surreal reciprocal. Show that the product of 𝐴 and a left element is less than one and the product of 𝐴 and a right element is greater than one. (Contributed by Scott Fenton, 14-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐴)    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))( 0s <s 𝑥𝑂 → ∃𝑦 ∈ No (𝑥𝑂 ·s 𝑦) = 1s ))    ⇒   ⊢ ((𝜑 ∧ 𝐼 ∈ ω) → (∀𝑏 ∈ (𝐿‘𝐼)(𝐴 ·s 𝑏) <s 1s ∧ ∀𝑐 ∈ (𝑅‘𝐼) 1s <s (𝐴 ·s 𝑐)))
Theorem	precsexlem10 28197*	Lemma for surreal reciprocal. Show that the union of the left sets is less than the union of the right sets. Note that this is the first theorem in the surreal numbers to require the axiom of infinity. (Contributed by Scott Fenton, 15-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐴)    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))( 0s <s 𝑥𝑂 → ∃𝑦 ∈ No (𝑥𝑂 ·s 𝑦) = 1s ))    ⇒   ⊢ (𝜑 → ∪ (𝐿 “ ω) <<s ∪ (𝑅 “ ω))
Theorem	precsexlem11 28198*	Lemma for surreal reciprocal. Show that the cut of the left and right sets is a multiplicative inverse for 𝐴. (Contributed by Scott Fenton, 15-Mar-2025.)
⊢ 𝐹 = rec((𝑝 ∈ V ↦ ⦋(1st ‘𝑝) / 𝑙⦌⦋(2nd ‘𝑝) / 𝑟⦌⟨(𝑙 ∪ ({𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝑅)} ∪ {𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝐿)})), (𝑟 ∪ ({𝑎 ∣ ∃𝑥𝐿 ∈ {𝑥 ∈ ( L ‘𝐴) ∣ 0s <s 𝑥}∃𝑦𝐿 ∈ 𝑙 𝑎 = (( 1s +s ((𝑥𝐿 -s 𝐴) ·s 𝑦𝐿)) /su 𝑥𝐿)} ∪ {𝑎 ∣ ∃𝑥𝑅 ∈ ( R ‘𝐴)∃𝑦𝑅 ∈ 𝑟 𝑎 = (( 1s +s ((𝑥𝑅 -s 𝐴) ·s 𝑦𝑅)) /su 𝑥𝑅)}))⟩), ⟨{ 0s }, ∅⟩)    &   ⊢ 𝐿 = (1st ∘ 𝐹)    &   ⊢ 𝑅 = (2nd ∘ 𝐹)    &   ⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐴)    &   ⊢ (𝜑 → ∀𝑥𝑂 ∈ (( L ‘𝐴) ∪ ( R ‘𝐴))( 0s <s 𝑥𝑂 → ∃𝑦 ∈ No (𝑥𝑂 ·s 𝑦) = 1s ))    &   ⊢ 𝑌 = (∪ (𝐿 “ ω) |s ∪ (𝑅 “ ω))    ⇒   ⊢ (𝜑 → (𝐴 ·s 𝑌) = 1s )
Theorem	precsex 28199*	Every positive surreal has a reciprocal. Theorem 10(iv) of [Conway] p. 21. (Contributed by Scott Fenton, 15-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 0s <s 𝐴) → ∃𝑦 ∈ No (𝐴 ·s 𝑦) = 1s )
Theorem	recsex 28200*	A non-zero surreal has a reciprocal. (Contributed by Scott Fenton, 15-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 𝐴 ≠ 0s ) → ∃𝑥 ∈ No (𝐴 ·s 𝑥) = 1s )