P. 285 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 28401-28500   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	mulscld 28401	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	ltmuls 28402	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	ltmulsd 28403	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	lemulsd 28404	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 28405	Surreal multiplication is commutative. 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 28406	Surreal multiplication is commutative. 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 28407	Surreal multiplication by zero. (Contributed by Scott Fenton, 4-Feb-2025.)
⊢ (𝐴 ∈ No → ( 0s ·s 𝐴) = 0s )
Theorem	mulslid 28408	Surreal one is a left identity element for multiplication. (Contributed by Scott Fenton, 4-Feb-2025.)
⊢ (𝐴 ∈ No → ( 1s ·s 𝐴) = 𝐴)
Theorem	mulslidd 28409	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 28410	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 28411	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 28412	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	sltmuls1 28413*	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	sltmuls2 28414*	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 28415*	Lemma for mulsunif 28416. 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 28416*	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 28417*	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 28418*	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 28419*	Lemma for addsdi 28421. 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 28420*	Lemma for addsdi 28421. 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 28421	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 28422	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 28423	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 28424	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 28425	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 28426	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 28427	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 28428	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 28429*	Lemma for mulsass 28432. 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 28430*	Lemma for mulsass 28432. 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 28431*	Lemma for mulsass 28432. 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 28432	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 28405, addsdi 28421, mulsgt0 28410, 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 28433	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 28434	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 28435*	Lemma for mulsunif2 28436. 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 28436*	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	ltmuls2 28437	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	ltmuls2d 28438	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	ltmuls1d 28439	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	lemuls2d 28440	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	lemuls1d 28441	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	ltmulnegs1d 28442	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	ltmulnegs2d 28443	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 28444	Lemma for mulscan2d 28445. 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 28445	Cancellation of surreal multiplication when the right term is nonzero. (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 28446	Cancellation of surreal multiplication when the left term is nonzero. (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 28447	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	lemuls1ad 28448	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	ltmuls12ad 28449	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 28450*	Uniqueness of surreal inversion. Given a nonzero 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 28451	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 28452	The product of two nonzero surreals is nonzero. (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 28453	Declare the syntax for surreal division.
class /su
Definition	df-divs 28454*	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 28455*	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 28456*	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 28457*	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 28458*	If a surreal has a reciprocal, then it is nonzero. (Contributed by Scott Fenton, 5-Sep-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐴 ·s 𝑥) = 1s )    ⇒   ⊢ (𝜑 → 𝐴 ≠ 0s )
Theorem	divmulsw 28459*	Relationship between surreal division and multiplication. Weak version that does not assume reciprocals. Later, when we prove precsex 28484, we can eliminate the existence hypothesis (see divmuls 28487). (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	divmulswd 28460*	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 28461*	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 28462*	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 28463*	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 28464*	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	ltdivmulswd 28465*	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	ltdivmuls2wd 28466*	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	ltmuldivswd 28467*	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	ltmuldivs2wd 28468*	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 28469*	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 28470	A surreal divided by one is itself. (Contributed by Scott Fenton, 13-Mar-2025.)
⊢ (𝐴 ∈ No → (𝐴 /su 1s ) = 𝐴)
Theorem	divs1d 28471	A surreal divided by one is itself. Deduction version. (Contributed by Scott Fenton, 27-Feb-2026.)
⊢ (𝜑 → 𝐴 ∈ No )    ⇒   ⊢ (𝜑 → (𝐴 /su 1s ) = 𝐴)
Theorem	precsexlemcbv 28472*	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 28473	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 28474	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 28475*	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 28476*	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 28477*	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 28478*	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 28479*	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 28480*	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 28481*	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 28482*	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 28483*	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 28484*	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 28485*	A nonzero surreal has a reciprocal. (Contributed by Scott Fenton, 15-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 𝐴 ≠ 0s ) → ∃𝑥 ∈ No (𝐴 ·s 𝑥) = 1s )
Theorem	recsexd 28486*	A nonzero surreal has a reciprocal. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐴 ≠ 0s )    ⇒   ⊢ (𝜑 → ∃𝑥 ∈ No (𝐴 ·s 𝑥) = 1s )
Theorem	divmuls 28487	Relationship between surreal division and multiplication. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ (𝐶 ∈ No ∧ 𝐶 ≠ 0s )) → ((𝐴 /su 𝐶) = 𝐵 ↔ (𝐶 ·s 𝐵) = 𝐴))
Theorem	divmulsd 28488	Relationship between surreal division and multiplication. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐶 ≠ 0s )    ⇒   ⊢ (𝜑 → ((𝐴 /su 𝐶) = 𝐵 ↔ (𝐶 ·s 𝐵) = 𝐴))
Theorem	divscl 28489	Surreal division closure law. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) → (𝐴 /su 𝐵) ∈ No )
Theorem	divscld 28490	Surreal division closure law. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐵 ≠ 0s )    ⇒   ⊢ (𝜑 → (𝐴 /su 𝐵) ∈ No )
Theorem	divscan2d 28491	A cancellation law for surreal division. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐵 ≠ 0s )    ⇒   ⊢ (𝜑 → (𝐵 ·s (𝐴 /su 𝐵)) = 𝐴)
Theorem	divscan1d 28492	A cancellation law for surreal division. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐵 ≠ 0s )    ⇒   ⊢ (𝜑 → ((𝐴 /su 𝐵) ·s 𝐵) = 𝐴)
Theorem	ltdivmulsd 28493	Surreal less-than relationship between division and multiplication. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    ⇒   ⊢ (𝜑 → ((𝐴 /su 𝐶) <s 𝐵 ↔ 𝐴 <s (𝐶 ·s 𝐵)))
Theorem	ltdivmuls2d 28494	Surreal less-than relationship between division and multiplication. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    ⇒   ⊢ (𝜑 → ((𝐴 /su 𝐶) <s 𝐵 ↔ 𝐴 <s (𝐵 ·s 𝐶)))
Theorem	ltmuldivsd 28495	Surreal less-than relationship between division and multiplication. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐶) <s 𝐵 ↔ 𝐴 <s (𝐵 /su 𝐶)))
Theorem	ltmuldivs2d 28496	Surreal less-than relationship between division and multiplication. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 0s <s 𝐶)    ⇒   ⊢ (𝜑 → ((𝐶 ·s 𝐴) <s 𝐵 ↔ 𝐴 <s (𝐵 /su 𝐶)))
Theorem	divsassd 28497	An associative law for surreal division. (Contributed by Scott Fenton, 16-Mar-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐶 ≠ 0s )    ⇒   ⊢ (𝜑 → ((𝐴 ·s 𝐵) /su 𝐶) = (𝐴 ·s (𝐵 /su 𝐶)))
Theorem	divmuldivsd 28498	Multiplication of two surreal ratios. (Contributed by Scott Fenton, 16-Apr-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐷 ∈ No )    &   ⊢ (𝜑 → 𝐵 ≠ 0s )    &   ⊢ (𝜑 → 𝐷 ≠ 0s )    ⇒   ⊢ (𝜑 → ((𝐴 /su 𝐵) ·s (𝐶 /su 𝐷)) = ((𝐴 ·s 𝐶) /su (𝐵 ·s 𝐷)))
Theorem	divdivs1d 28499	Surreal division into a fraction. (Contributed by Scott Fenton, 7-Aug-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐶 ∈ No )    &   ⊢ (𝜑 → 𝐵 ≠ 0s )    &   ⊢ (𝜑 → 𝐶 ≠ 0s )    ⇒   ⊢ (𝜑 → ((𝐴 /su 𝐵) /su 𝐶) = (𝐴 /su (𝐵 ·s 𝐶)))
Theorem	divsrecd 28500	Relationship between surreal division and reciprocal. (Contributed by Scott Fenton, 13-Aug-2025.)
⊢ (𝜑 → 𝐴 ∈ No )    &   ⊢ (𝜑 → 𝐵 ∈ No )    &   ⊢ (𝜑 → 𝐵 ≠ 0s )    ⇒   ⊢ (𝜑 → (𝐴 /su 𝐵) = (𝐴 ·s ( 1s /su 𝐵)))