P. 281 of Theorem List - Metamath Proof Explorer

Theorem List for Metamath Proof Explorer - 28001-28100   *Has distinct variable group(s)

Type	Label	Description
Statement
Theorem	fltabcoprmex 28001	A counterexample to FLT implies a counterexample to FLT with 𝐴, 𝐵 (assigned to 𝐴 / (𝐴 gcd 𝐵) and 𝐵 / (𝐴 gcd 𝐵)) coprime (by divgcdcoprm0 16869). (Contributed by SN, 20-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    ⇒   ⊢ (𝜑 → (((𝐴 / (𝐴 gcd 𝐵))↑𝑁) + ((𝐵 / (𝐴 gcd 𝐵))↑𝑁)) = ((𝐶 / (𝐴 gcd 𝐵))↑𝑁))
Theorem	fltaccoprm 28002	A counterexample to FLT with 𝐴, 𝐵 coprime also has 𝐴, 𝐶 coprime. (Contributed by SN, 20-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    ⇒   ⊢ (𝜑 → (𝐴 gcd 𝐶) = 1)
Theorem	fltbccoprm 28003	A counterexample to FLT with 𝐴, 𝐵 coprime also has 𝐵, 𝐶 coprime. Proven from fltaccoprm 28002 using commutativity of addition. (Contributed by SN, 20-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ ℕ0)    &   ⊢ (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    ⇒   ⊢ (𝜑 → (𝐵 gcd 𝐶) = 1)
Theorem	fltabcoprm 28004	A counterexample to FLT with 𝐴, 𝐶 coprime also has 𝐴, 𝐵 coprime. Converse of fltaccoprm 28002. (Contributed by SN, 22-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → (𝐴 gcd 𝐶) = 1)    &   ⊢ (𝜑 → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)
Theorem	fltne 28005	If a counterexample to FLT exists, its addends are not equal. (Contributed by SN, 1-Jun-2023.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘2))    &   ⊢ (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) = (𝐶↑𝑁))    ⇒   ⊢ (𝜑 → 𝐴 ≠ 𝐵)
Theorem	flt4lem1 28006	Satisfy the antecedent used in several pythagtrip 17041 lemmas. (Contributed by SN, 21-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝐴)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → ((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)))
Theorem	flt4lem2 28007	If 𝐴 is even, 𝐵 is odd. (Contributed by SN, 22-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 2 ∥ 𝐴)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    ⇒   ⊢ (𝜑 → ¬ 2 ∥ 𝐵)
Theorem	flt4lem3 28008	Equivalent to pythagtriplem4 17026. Show that 𝐶 + 𝐴 and 𝐶 − 𝐴 are coprime. (Contributed by SN, 22-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → 2 ∥ 𝐴)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → ((𝐶 + 𝐴) gcd (𝐶 − 𝐴)) = 1)
Theorem	flt4lem4 28009	If the product of two coprime factors is a perfect square, the factors are perfect squares. (Contributed by SN, 22-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → (𝐴 · 𝐵) = (𝐶↑2))    ⇒   ⊢ (𝜑 → (𝐴 = ((𝐴 gcd 𝐶)↑2) ∧ 𝐵 = ((𝐵 gcd 𝐶)↑2)))
Theorem	flt4lem5 28010	In the context of the lemmas of pythagtrip 17041, 𝑀 and 𝑁 are coprime. (Contributed by SN, 23-Aug-2024.)
⊢ 𝑀 = (((√‘(𝐶 + 𝐵)) + (√‘(𝐶 − 𝐵))) / 2)    &   ⊢ 𝑁 = (((√‘(𝐶 + 𝐵)) − (√‘(𝐶 − 𝐵))) / 2)    ⇒   ⊢ (((𝐴 ∈ ℕ ∧ 𝐵 ∈ ℕ ∧ 𝐶 ∈ ℕ) ∧ ((𝐴↑2) + (𝐵↑2)) = (𝐶↑2) ∧ ((𝐴 gcd 𝐵) = 1 ∧ ¬ 2 ∥ 𝐴)) → (𝑀 gcd 𝑁) = 1)
Theorem	flt4lem5elem 28011	Version of fltaccoprm 28002 and fltbccoprm 28003 where 𝑀 is not squared. This can be proved in general for any polynomial in three variables: using prmdvdsncoprmbd 16932, dvds2addd 16491, and prmdvdsexp 16920, we can show that if two variables are coprime, the third is also coprime to the two. (Contributed by SN, 24-Aug-2024.)
⊢ (𝜑 → 𝑀 ∈ ℕ)    &   ⊢ (𝜑 → 𝑅 ∈ ℕ)    &   ⊢ (𝜑 → 𝑆 ∈ ℕ)    &   ⊢ (𝜑 → 𝑀 = ((𝑅↑2) + (𝑆↑2)))    &   ⊢ (𝜑 → (𝑅 gcd 𝑆) = 1)    ⇒   ⊢ (𝜑 → ((𝑅 gcd 𝑀) = 1 ∧ (𝑆 gcd 𝑀) = 1))
Theorem	flt4lem5a 28012	Part 1 of Equation 1 of https://crypto.stanford.edu/pbc/notes/numberfield/fermatn4.html. (Contributed by SN, 22-Aug-2024.)
⊢ 𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ 𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝐴)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → ((𝐴↑2) + (𝑁↑2)) = (𝑀↑2))
Theorem	flt4lem5b 28013	Part 2 of Equation 1 of https://crypto.stanford.edu/pbc/notes/numberfield/fermatn4.html. (Contributed by SN, 22-Aug-2024.)
⊢ 𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ 𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝐴)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → (2 · (𝑀 · 𝑁)) = (𝐵↑2))
Theorem	flt4lem5c 28014	Part 2 of Equation 2 of https://crypto.stanford.edu/pbc/notes/numberfield/fermatn4.html. (Contributed by SN, 22-Aug-2024.)
⊢ 𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ 𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝐴)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → 𝑁 = (2 · (𝑅 · 𝑆)))
Theorem	flt4lem5d 28015	Part 3 of Equation 2 of https://crypto.stanford.edu/pbc/notes/numberfield/fermatn4.html. (Contributed by SN, 23-Aug-2024.)
⊢ 𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ 𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝐴)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → 𝑀 = ((𝑅↑2) + (𝑆↑2)))
Theorem	flt4lem5e 28016	Satisfy the hypotheses of flt4lem4 28009. (Contributed by SN, 23-Aug-2024.)
⊢ 𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ 𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝐴)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → (((𝑅 gcd 𝑆) = 1 ∧ (𝑅 gcd 𝑀) = 1 ∧ (𝑆 gcd 𝑀) = 1) ∧ (𝑅 ∈ ℕ ∧ 𝑆 ∈ ℕ ∧ 𝑀 ∈ ℕ) ∧ ((𝑀 · (𝑅 · 𝑆)) = ((𝐵 / 2)↑2) ∧ (𝐵 / 2) ∈ ℕ)))
Theorem	flt4lem5f 28017	Final equation of https://crypto.stanford.edu/pbc/notes/numberfield/fermatn4.html. Given 𝐴↑4 + 𝐵↑4 = 𝐶↑2, provide a smaller solution. This satisfies the infinite descent condition. (Contributed by SN, 24-Aug-2024.)
⊢ 𝑀 = (((√‘(𝐶 + (𝐵↑2))) + (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑁 = (((√‘(𝐶 + (𝐵↑2))) − (√‘(𝐶 − (𝐵↑2)))) / 2)    &   ⊢ 𝑅 = (((√‘(𝑀 + 𝑁)) + (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ 𝑆 = (((√‘(𝑀 + 𝑁)) − (√‘(𝑀 − 𝑁))) / 2)    &   ⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝐴)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → ((𝑀 gcd (𝐵 / 2))↑2) = (((𝑅 gcd (𝐵 / 2))↑4) + ((𝑆 gcd (𝐵 / 2))↑4)))
Theorem	flt4lem6 28018	Remove shared factors in a solution to 𝐴↑4 + 𝐵↑4 = 𝐶↑2. (Contributed by SN, 24-Jul-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → (((𝐴 / (𝐴 gcd 𝐵)) ∈ ℕ ∧ (𝐵 / (𝐴 gcd 𝐵)) ∈ ℕ ∧ (𝐶 / ((𝐴 gcd 𝐵)↑2)) ∈ ℕ) ∧ (((𝐴 / (𝐴 gcd 𝐵))↑4) + ((𝐵 / (𝐴 gcd 𝐵))↑4)) = ((𝐶 / ((𝐴 gcd 𝐵)↑2))↑2)))
Theorem	flt4lem7 28019*	Convert flt4lem5f 28017 into a convenient form for nna4b4nsq 28020. (Contributed by SN, 25-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → ¬ 2 ∥ 𝐴)    &   ⊢ (𝜑 → (𝐴 gcd 𝐵) = 1)    &   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) = (𝐶↑2))    ⇒   ⊢ (𝜑 → ∃𝑙 ∈ ℕ (∃𝑔 ∈ ℕ ∃ℎ ∈ ℕ (¬ 2 ∥ 𝑔 ∧ ((𝑔 gcd ℎ) = 1 ∧ ((𝑔↑4) + (ℎ↑4)) = (𝑙↑2))) ∧ 𝑙 < 𝐶))
Theorem	nna4b4nsq 28020	Strengthening of Fermat's last theorem for exponent 4, where the sum is only assumed to be a square. (Contributed by SN, 23-Aug-2024.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    ⇒   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) ≠ (𝐶↑2))
Theorem	flt4 28021	Fermat's last theorem for the exponent four. (Contributed by AV, 15-Sep-2026.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    ⇒   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) ≠ (𝐶↑4))
Theorem	flt4ALT 28022*	Fermat's last theorem for the exponent four, derived from Fermat's right triangle theorem (which is not proved yet, see fermrtt: after a proof is available, the hypothesis flt4ALT.r can be removed - TODO-AV). (Contributed by AV, 15-Sep-2026.) (Proof modification is discouraged.) (New usage is discouraged.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → ∀𝑎 ∈ ℕ ∀𝑏 ∈ ℕ ∀𝑐 ∈ ℕ ((𝑎↑4) − (𝑏↑4)) ≠ (𝑐↑2))    ⇒   ⊢ (𝜑 → ((𝐴↑4) + (𝐵↑4)) ≠ (𝐶↑4))
14.4.15.2  Proof of Fermat's Last Theorem (work in progress) On September 4th, 2026 Anthropic published that their AI "Claude" created a complete and machine-verified formal proof for Fermat's Last Theorem: it consists of about 13 million lines of code, and about 30.000 verified theorems, created within about 11 days, built on Lean's library "mathlib" as well as preliminary work of the Buzzard community (in 2024, Kevin Buzzard started a project at Imperial College to formalize a complete proof in Lean, planned for many years and founded up to 2029). For more details see https://www.anthropic.com/research/formalizing-fermats-last-theorem. Asking Claude (Opus 5) itself if it is possible to prove Fermat's last theorem also with Metamath, it answered: "Short version: in principle yes, in practice no. Foundations aren't the problem. set.mm is built on ZFC, and in the Grothendieck extension (ax-groth 10935, Tarski's axiom) universes are available too - exactly what Wiles's proof appeared to need beyond ZFC. (Colin McLarty has shown in any case that this can be pushed down into far weaker systems.) So there is no logical obstacle, only a labor one. What has been proved so far is the case n = 4 (see flt4 28021). ... Why it's nevertheless a size too big: Metamath has no tactics; every step stands there fully expanded. More serious is that set.mm simply lacks the necessary infrastructure: no schemes, no modular forms, no Galois representations, no class field theory. That is the actual proof - the number theory of FLT is only the last line. Mathlib had a large part of it after years of community work, and even building on that, the Lean formalization comprised some 30,000 additional theorems; set.mm in total holds only an order of magnitude more than that. ... So it would be feasible - but it would be a project on the same scale as everything assembled in set.mm since 1993." Nevertheless, this and the previous subsections provide some starting points for a formal proof by Metamath: According to the blueprint provided by Kevin Buzzard, see https://imperialcollegelondon.github.io/FLT/blueprint/ 28021, Lemma 2.1 is already proven (see fltoprm 28025). The next step would be to prove Lemma 2.2 (FLT for n = 3, see flt3 above). Corollary 2.3 follows immediately from Lemmas 2.1 and 2.2 (see fltoprmgt3 28026). Finally, asking Claude if it makes sense to translate the Lean proof into the Metamath language, it says: "Possible in principle - sensible, not really. Foundationally there's no obstacle. Mario Carneiro showed that Lean's type theory is consistent relative to ZFC with n inaccessible cardinals for all n < ω. set.mm, via Tarski's axiom (ax-groth 10935), supplies arbitrarily large inaccessibles - so the required model is within reach. And there's precedent: in 2014/16 Carneiro translated proofs from the OpenTheory format (HOL Light, HOL4, Isabelle) into Metamath, first into a HOL formalization hol.mm, then embedded into set.mm. The exercise has been done once already. But Lean is a different league from HOL. Three problems: * The translation isn't a rewrite. ... * The transfer problem. ... * The scale. Lean's elaborator performs an enormous amount of invisible work: instance resolution, implicit arguments, universe unification, definitional unfolding. Metamath has none of it. 13 million lines of Lean on top of over a million lines of mathlib plausibly expand into the billions of steps - against the few tens of thousands of statements that make up all of set.mm. The real objection, though, isn't feasibility but payoff. ... Nothing for set.mm. The database is curated and maintained for readability; a generated dump would contradict its purpose and wouldn't be admitted into the main database anyway. What does transfer is the blueprint. The lemma decomposition, the definitional choices, the proof strategy - that's language-independent, and it's precisely the part that took years of work to produce. Buzzard's blueprint for the FLT project is literally that document. A native Metamath development following it would be the sensible path, and it would reasonably start at n = 3.". The last sentence is actually the way we want to continue (as described above)...
Theorem	fltoprmlem1 28023*	Lemma 1 for fltoprm 28025: Every positive integer is either a power of 2 or has an odd prime factor. (Contributed by AV, 15-Sep-2026.)
⊢ (𝑁 ∈ ℕ → (∃𝑝 ∈ ℙ (2 < 𝑝 ∧ 𝑝 ∥ 𝑁) ∨ ∃𝑛 ∈ ℕ0 𝑁 = (2↑𝑛)))
Theorem	fltoprmlem2 28024	Lemma 2 for fltoprm 28025: If an integer greater than or equal to 3 is a power of 2, it is a multiple of 4. (Contributed by AV, 15-Sep-2026.)
⊢ ((𝑁 ∈ (ℤ≥‘3) ∧ 𝐾 ∈ ℕ0 ∧ 𝑁 = (2↑𝐾)) → 4 ∥ 𝑁)
Theorem	fltoprm 28025*	Fermat's last theorem holds for any exponent greater than 2 if it holds for odd prime exponents. (Contributed by AV, 15-Sep-2026.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘3))    &   ⊢ (𝜑 → ∀𝑎 ∈ ℕ ∀𝑏 ∈ ℕ ∀𝑐 ∈ ℕ ∀𝑝 ∈ ℙ (2 < 𝑝 → ((𝑎↑𝑝) + (𝑏↑𝑝)) ≠ (𝑐↑𝑝)))    ⇒   ⊢ (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) ≠ (𝐶↑𝑁))
Theorem	fltoprmgt3 28026*	Fermat's last theorem holds for any exponent greater than 2 if it holds for all odd prime exponents greater than 3. (TODO-AV: after a proof is available for 𝑁 = 3, see flt3, the hypothesis fltoprmgt3.3 can be removed.) (Contributed by AV, 15-Sep-2026.)
⊢ (𝜑 → 𝐴 ∈ ℕ)    &   ⊢ (𝜑 → 𝐵 ∈ ℕ)    &   ⊢ (𝜑 → 𝐶 ∈ ℕ)    &   ⊢ (𝜑 → 𝑁 ∈ (ℤ≥‘3))    &   ⊢ (𝜑 → ∀𝑎 ∈ ℕ ∀𝑏 ∈ ℕ ∀𝑐 ∈ ℕ ∀𝑝 ∈ ℙ (3 < 𝑝 → ((𝑎↑𝑝) + (𝑏↑𝑝)) ≠ (𝑐↑𝑝)))    &   ⊢ (𝜑 → ((𝑎↑3) + (𝑏↑3)) ≠ (𝑐↑3))    ⇒   ⊢ (𝜑 → ((𝐴↑𝑁) + (𝐵↑𝑁)) ≠ (𝐶↑𝑁))
PART 15  SURREAL NUMBERS
15.1  Sign sequence representation and Alling's axioms The surreal numbers can be represented in several equivalent ways. In [Alling], Norman Alling made this notion explicit by giving a set of axioms that all representations admit, then proving that there is an order and birthday preserving bijection between any systems that satisfy these axioms. In this section, we start with the definition of surreal numbers given in [Gonshor] and derive Alling's axioms. After deriving them we no longer refer to the explicit definition of surreals. In particular, we never take advantage of the fact that the empty set is a surreal number under our definition.
15.1.1  Definitions and initial properties
Syntax	csur 28027	Declare the class of all surreal numbers (see df-no 28030).
class No
Syntax	clts 28028	Declare the less-than relation over surreal numbers (see df-lts 28031).
class <s
Syntax	cbday 28029	Declare the birthday function for surreal numbers (see df-bday 28032).
class bday
Definition	df-no 28030*	Define the class of surreal numbers. The surreal numbers are a proper class of numbers developed by John H. Conway and introduced by Donald Knuth in 1975. They form a proper class into which all ordered fields can be embedded. The approach we take to defining them was first introduced by Hary Gonshor, and is based on the conception of a "sign expansion" of a surreal number. We define the surreals as ordinal-indexed sequences of 1o and 2o, analogous to Gonshor's ( − ) and ( + ). After introducing this definition, we will abstract away from it using axioms that Norman Alling developed in "Foundations of Analysis over Surreal Number Fields." This is done in an effort to be agnostic towards the exact implementation of surreals. (Contributed by Scott Fenton, 9-Jun-2011.)
⊢ No = {𝑓 ∣ ∃𝑎 ∈ On 𝑓:𝑎⟶{1o, 2o}}
Definition	df-lts 28031*	Next, we introduce surreal less-than, a comparison relation over the surreals by lexicographically ordering them. (Contributed by Scott Fenton, 9-Jun-2011.)
⊢ <s = {⟨𝑓, 𝑔⟩ ∣ ((𝑓 ∈ No ∧ 𝑔 ∈ No) ∧ ∃𝑥 ∈ On (∀𝑦 ∈ 𝑥 (𝑓‘𝑦) = (𝑔‘𝑦) ∧ (𝑓‘𝑥){⟨1o, ∅⟩, ⟨1o, 2o⟩, ⟨∅, 2o⟩} (𝑔‘𝑥)))}
Definition	df-bday 28032	Finally, we introduce the birthday function. This function maps each surreal to an ordinal. In our implementation, this is the domain of the sign function. The important properties of this function are established later. (Contributed by Scott Fenton, 11-Jun-2011.)
⊢ bday = (𝑥 ∈ No ↦ dom 𝑥)
Theorem	elno 28033*	Membership in the surreals. (Contributed by Scott Fenton, 11-Jun-2011.) (Proof shortened by SF, 14-Apr-2012.) Avoid ax-rep 5235. (Revised by SN, 5-Jun-2025.)
⊢ (𝐴 ∈ No ↔ ∃𝑥 ∈ On 𝐴:𝑥⟶{1o, 2o})
Theorem	ltsval 28034*	The value of the surreal less-than relation. (Contributed by Scott Fenton, 14-Jun-2011.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No) → (𝐴 <s 𝐵 ↔ ∃𝑥 ∈ On (∀𝑦 ∈ 𝑥 (𝐴‘𝑦) = (𝐵‘𝑦) ∧ (𝐴‘𝑥){⟨1o, ∅⟩, ⟨1o, 2o⟩, ⟨∅, 2o⟩} (𝐵‘𝑥))))
Theorem	bdayval 28035	The value of the birthday function within the surreals. (Contributed by Scott Fenton, 14-Jun-2011.)
⊢ (𝐴 ∈ No → (bday‘𝐴) = dom 𝐴)
Theorem	nofun 28036	A surreal is a function. (Contributed by Scott Fenton, 16-Jun-2011.)
⊢ (𝐴 ∈ No → Fun 𝐴)
Theorem	nodmon 28037	The domain of a surreal is an ordinal. (Contributed by Scott Fenton, 16-Jun-2011.)
⊢ (𝐴 ∈ No → dom 𝐴 ∈ On)
Theorem	norn 28038	The range of a surreal is a subset of the surreal signs. (Contributed by Scott Fenton, 16-Jun-2011.)
⊢ (𝐴 ∈ No → ran 𝐴 ⊆ {1o, 2o})
Theorem	nofnbday 28039	A surreal is a function over its birthday. (Contributed by Scott Fenton, 16-Jun-2011.)
⊢ (𝐴 ∈ No → 𝐴 Fn (bday‘𝐴))
Theorem	nodmord 28040	The domain of a surreal has the ordinal property. (Contributed by Scott Fenton, 16-Jun-2011.)
⊢ (𝐴 ∈ No → Ord dom 𝐴)
Theorem	elno2 28041	An alternative condition for membership in No. (Contributed by Scott Fenton, 21-Mar-2012.)
⊢ (𝐴 ∈ No ↔ (Fun 𝐴 ∧ dom 𝐴 ∈ On ∧ ran 𝐴 ⊆ {1o, 2o}))
Theorem	elno3 28042	Another condition for membership in No. (Contributed by Scott Fenton, 14-Apr-2012.)
⊢ (𝐴 ∈ No ↔ (𝐴:dom 𝐴⟶{1o, 2o} ∧ dom 𝐴 ∈ On))
Theorem	ltsval2 28043*	Alternate expression for surreal less-than. Two surreals obey surreal less-than iff they obey the sign ordering at the first place they differ. (Contributed by Scott Fenton, 17-Jun-2011.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No) → (𝐴 <s 𝐵 ↔ (𝐴‘∩ {𝑎 ∈ On ∣ (𝐴‘𝑎) ≠ (𝐵‘𝑎)}){⟨1o, ∅⟩, ⟨1o, 2o⟩, ⟨∅, 2o⟩} (𝐵‘∩ {𝑎 ∈ On ∣ (𝐴‘𝑎) ≠ (𝐵‘𝑎)})))
Theorem	nofv 28044	The function value of a surreal is either a sign or the empty set. (Contributed by Scott Fenton, 22-Jun-2011.)
⊢ (𝐴 ∈ No → ((𝐴‘𝑋) = ∅ ∨ (𝐴‘𝑋) = 1o ∨ (𝐴‘𝑋) = 2o))
Theorem	nosgnn0 28045	∅ is not a surreal sign. (Contributed by Scott Fenton, 16-Jun-2011.)
⊢ ¬ ∅ ∈ {1o, 2o}
Theorem	nosgnn0i 28046	If 𝑋 is a surreal sign, then it is not null. (Contributed by Scott Fenton, 3-Aug-2011.)
⊢ 𝑋 ∈ {1o, 2o}    ⇒   ⊢ ∅ ≠ 𝑋
Theorem	noreson 28047	The restriction of a surreal to an ordinal is still a surreal. (Contributed by Scott Fenton, 4-Sep-2011.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ On) → (𝐴 ↾ 𝐵) ∈ No)
Theorem	ltsintdifex 28048*	If 𝐴 <s 𝐵, then the intersection of all the ordinals that have differing signs in 𝐴 and 𝐵 exists. (Contributed by Scott Fenton, 22-Feb-2012.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No) → (𝐴 <s 𝐵 → ∩ {𝑎 ∈ On ∣ (𝐴‘𝑎) ≠ (𝐵‘𝑎)} ∈ V))
Theorem	ltsres 28049	If the restrictions of two surreals to a given ordinal obey surreal less-than, then so do the two surreals themselves. (Contributed by Scott Fenton, 4-Sep-2011.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝑋 ∈ On) → ((𝐴 ↾ 𝑋) <s (𝐵 ↾ 𝑋) → 𝐴 <s 𝐵))
Theorem	noxp1o 28050	The Cartesian product of an ordinal and {1o} is a surreal. (Contributed by Scott Fenton, 12-Jun-2011.)
⊢ (𝐴 ∈ On → (𝐴 × {1o}) ∈ No)
Theorem	noseponlem 28051*	Lemma for nosepon 28052. Consider a case of proper subset domain. (Contributed by Scott Fenton, 21-Sep-2020.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ dom 𝐴 ∈ dom 𝐵) → ¬ ∀𝑥 ∈ On (𝐴‘𝑥) = (𝐵‘𝑥))
Theorem	nosepon 28052*	Given two unequal surreals, the minimal ordinal at which they differ is an ordinal. (Contributed by Scott Fenton, 21-Sep-2020.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐴 ≠ 𝐵) → ∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ (𝐵‘𝑥)} ∈ On)
Theorem	noextend 28053	Extending a surreal by one sign value results in a new surreal. (Contributed by Scott Fenton, 22-Nov-2021.)
⊢ 𝑋 ∈ {1o, 2o}    ⇒   ⊢ (𝐴 ∈ No → (𝐴 ∪ {⟨dom 𝐴, 𝑋⟩}) ∈ No)
Theorem	noextendseq 28054	Extend a surreal by a sequence of ordinals. (Contributed by Scott Fenton, 30-Nov-2021.)
⊢ 𝑋 ∈ {1o, 2o}    ⇒   ⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ On) → (𝐴 ∪ ((𝐵 ∖ dom 𝐴) × {𝑋})) ∈ No)
Theorem	noextenddif 28055*	Calculate the place where a surreal and its extension differ. (Contributed by Scott Fenton, 22-Nov-2021.)
⊢ 𝑋 ∈ {1o, 2o}    ⇒   ⊢ (𝐴 ∈ No → ∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ ((𝐴 ∪ {⟨dom 𝐴, 𝑋⟩})‘𝑥)} = dom 𝐴)
Theorem	noextendlt 28056	Extending a surreal with a negative sign results in a smaller surreal. (Contributed by Scott Fenton, 22-Nov-2021.)
⊢ (𝐴 ∈ No → (𝐴 ∪ {⟨dom 𝐴, 1o⟩}) <s 𝐴)
Theorem	noextendgt 28057	Extending a surreal with a positive sign results in a bigger surreal. (Contributed by Scott Fenton, 22-Nov-2021.)
⊢ (𝐴 ∈ No → 𝐴 <s (𝐴 ∪ {⟨dom 𝐴, 2o⟩}))
Theorem	nolesgn2o 28058	Given 𝐴 less-than or equal to 𝐵, equal to 𝐵 up to 𝑋, and 𝐴(𝑋) = 2o, then 𝐵(𝑋) = 2o. (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝑋 ∈ On) ∧ ((𝐴 ↾ 𝑋) = (𝐵 ↾ 𝑋) ∧ (𝐴‘𝑋) = 2o) ∧ ¬ 𝐵 <s 𝐴) → (𝐵‘𝑋) = 2o)
Theorem	nolesgn2ores 28059	Given 𝐴 less-than or equal to 𝐵, equal to 𝐵 up to 𝑋, and 𝐴(𝑋) = 2o, then (𝐴 ↾ suc 𝑋) = (𝐵 ↾ suc 𝑋). (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝑋 ∈ On) ∧ ((𝐴 ↾ 𝑋) = (𝐵 ↾ 𝑋) ∧ (𝐴‘𝑋) = 2o) ∧ ¬ 𝐵 <s 𝐴) → (𝐴 ↾ suc 𝑋) = (𝐵 ↾ suc 𝑋))
Theorem	nogesgn1o 28060	Given 𝐴 greater than or equal to 𝐵, equal to 𝐵 up to 𝑋, and 𝐴(𝑋) = 1o, then 𝐵(𝑋) = 1o. (Contributed by Scott Fenton, 9-Aug-2024.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝑋 ∈ On) ∧ ((𝐴 ↾ 𝑋) = (𝐵 ↾ 𝑋) ∧ (𝐴‘𝑋) = 1o) ∧ ¬ 𝐴 <s 𝐵) → (𝐵‘𝑋) = 1o)
Theorem	nogesgn1ores 28061	Given 𝐴 greater than or equal to 𝐵, equal to 𝐵 up to 𝑋, and 𝐴(𝑋) = 1o, then (𝐴 ↾ suc 𝑋) = (𝐵 ↾ suc 𝑋). (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝑋 ∈ On) ∧ ((𝐴 ↾ 𝑋) = (𝐵 ↾ 𝑋) ∧ (𝐴‘𝑋) = 1o) ∧ ¬ 𝐴 <s 𝐵) → (𝐴 ↾ suc 𝑋) = (𝐵 ↾ suc 𝑋))
15.1.2  Ordering
Theorem	ltssolem1 28062	Lemma for ltsso 28063. The "sign expansion" binary relation totally orders the surreal signs. (Contributed by Scott Fenton, 8-Jun-2011.)
⊢ {⟨1o, ∅⟩, ⟨1o, 2o⟩, ⟨∅, 2o⟩} Or ({1o, 2o} ∪ {∅})
Theorem	ltsso 28063	Less-than totally orders the surreals. Axiom O of [Alling] p. 184. (Contributed by Scott Fenton, 9-Jun-2011.)
⊢ <s Or No
15.1.3  Birthday Function
Theorem	bdayfo 28064	The birthday function maps the surreals onto the ordinals. Axiom B of [Alling] p. 184. (Proof shortened on 14-Apr-2012 by SF). (Contributed by Scott Fenton, 11-Jun-2011.)
⊢ bday:No⟶ontoOn
15.1.4  Density
Theorem	fvnobday 28065	The value of a surreal at its birthday is ∅. (Contributed by Scott Fenton, 14-Jun-2011.) (Proof shortened by SF, 14-Apr-2012.)
⊢ (𝐴 ∈ No → (𝐴‘(bday‘𝐴)) = ∅)
Theorem	nosepnelem 28066*	Lemma for nosepne 28067. (Contributed by Scott Fenton, 24-Nov-2021.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐴 <s 𝐵) → (𝐴‘∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ (𝐵‘𝑥)}) ≠ (𝐵‘∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ (𝐵‘𝑥)}))
Theorem	nosepne 28067*	The value of two non-equal surreals at the first place they differ is different. (Contributed by Scott Fenton, 24-Nov-2021.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐴 ≠ 𝐵) → (𝐴‘∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ (𝐵‘𝑥)}) ≠ (𝐵‘∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ (𝐵‘𝑥)}))
Theorem	nosep1o 28068*	If the value of a surreal at a separator is 1o then the surreal is lesser. (Contributed by Scott Fenton, 7-Dec-2021.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐴 ≠ 𝐵) ∧ (𝐴‘∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ (𝐵‘𝑥)}) = 1o) → 𝐴 <s 𝐵)
Theorem	nosep2o 28069*	If the value of a surreal at a separator is 2o then the surreal is greater. (Contributed by Scott Fenton, 7-Dec-2021.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐴 ≠ 𝐵) ∧ (𝐴‘∩ {𝑥 ∈ On ∣ (𝐵‘𝑥) ≠ (𝐴‘𝑥)}) = 2o) → 𝐵 <s 𝐴)
Theorem	nosepdmlem 28070*	Lemma for nosepdm 28071. (Contributed by Scott Fenton, 24-Nov-2021.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐴 <s 𝐵) → ∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ (𝐵‘𝑥)} ∈ (dom 𝐴 ∪ dom 𝐵))
Theorem	nosepdm 28071*	The first place two surreals differ is an element of the larger of their domains. (Contributed by Scott Fenton, 24-Nov-2021.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐴 ≠ 𝐵) → ∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ (𝐵‘𝑥)} ∈ (dom 𝐴 ∪ dom 𝐵))
Theorem	nosepeq 28072*	The values of two surreals at a point less than their separators are equal. (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐴 ≠ 𝐵) ∧ 𝑋 ∈ ∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ (𝐵‘𝑥)}) → (𝐴‘𝑋) = (𝐵‘𝑋))
Theorem	nosepssdm 28073*	Given two non-equal surreals, their separator is less-than or equal to the domain of one of them. Part of Lemma 2.1.1 of [Lipparini] p. 3. (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐴 ≠ 𝐵) → ∩ {𝑥 ∈ On ∣ (𝐴‘𝑥) ≠ (𝐵‘𝑥)} ⊆ dom 𝐴)
Theorem	nodenselem4 28074*	Lemma for nodense 28079. Show that a particular abstraction is an ordinal. (Contributed by Scott Fenton, 16-Jun-2011.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No) ∧ 𝐴 <s 𝐵) → ∩ {𝑎 ∈ On ∣ (𝐴‘𝑎) ≠ (𝐵‘𝑎)} ∈ On)
Theorem	nodenselem5 28075*	Lemma for nodense 28079. If the birthdays of two distinct surreals are equal, then the ordinal from nodenselem4 28074 is an element of that birthday. (Contributed by Scott Fenton, 16-Jun-2011.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No) ∧ ((bday‘𝐴) = (bday‘𝐵) ∧ 𝐴 <s 𝐵)) → ∩ {𝑎 ∈ On ∣ (𝐴‘𝑎) ≠ (𝐵‘𝑎)} ∈ (bday‘𝐴))
Theorem	nodenselem6 28076*	The restriction of a surreal to the abstraction from nodenselem4 28074 is still a surreal. (Contributed by Scott Fenton, 16-Jun-2011.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No) ∧ ((bday‘𝐴) = (bday‘𝐵) ∧ 𝐴 <s 𝐵)) → (𝐴 ↾ ∩ {𝑎 ∈ On ∣ (𝐴‘𝑎) ≠ (𝐵‘𝑎)}) ∈ No)
Theorem	nodenselem7 28077*	Lemma for nodense 28079. 𝐴 and 𝐵 are equal at all elements of the abstraction. (Contributed by Scott Fenton, 17-Jun-2011.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No) ∧ ((bday‘𝐴) = (bday‘𝐵) ∧ 𝐴 <s 𝐵)) → (𝐶 ∈ ∩ {𝑎 ∈ On ∣ (𝐴‘𝑎) ≠ (𝐵‘𝑎)} → (𝐴‘𝐶) = (𝐵‘𝐶)))
Theorem	nodenselem8 28078*	Lemma for nodense 28079. Give a condition for surreal less-than when two surreals have the same birthday. (Contributed by Scott Fenton, 19-Jun-2011.)
⊢ ((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ (bday‘𝐴) = (bday‘𝐵)) → (𝐴 <s 𝐵 ↔ ((𝐴‘∩ {𝑎 ∈ On ∣ (𝐴‘𝑎) ≠ (𝐵‘𝑎)}) = 1o ∧ (𝐵‘∩ {𝑎 ∈ On ∣ (𝐴‘𝑎) ≠ (𝐵‘𝑎)}) = 2o)))
Theorem	nodense 28079*	Given two distinct surreals with the same birthday, there is an older surreal lying between the two of them. Axiom SD of [Alling] p. 184. (Contributed by Scott Fenton, 16-Jun-2011.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No) ∧ ((bday‘𝐴) = (bday‘𝐵) ∧ 𝐴 <s 𝐵)) → ∃𝑥 ∈ No ((bday‘𝑥) ∈ (bday‘𝐴) ∧ 𝐴 <s 𝑥 ∧ 𝑥 <s 𝐵))
15.1.5  Full-Eta Property The theorems in this section are derived from "A clean way to separate sets of surreals" by Paolo Lipparini, https://doi.org/10.48550/arXiv.1712.03500.
Theorem	bdayimaon 28080	Lemma for full-eta properties. The successor of the union of the image of the birthday function under a set is an ordinal. (Contributed by Scott Fenton, 20-Aug-2011.)
⊢ (𝐴 ∈ 𝑉 → suc ∪ (bday “ 𝐴) ∈ On)
Theorem	nolt02olem 28081	Lemma for nolt02o 28082. If 𝐴(𝑋) is undefined with 𝐴 surreal and 𝑋 ordinal, then dom 𝐴 ⊆ 𝑋. (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ ((𝐴 ∈ No ∧ 𝑋 ∈ On ∧ (𝐴‘𝑋) = ∅) → dom 𝐴 ⊆ 𝑋)
Theorem	nolt02o 28082	Given 𝐴 less-than 𝐵, equal to 𝐵 up to 𝑋, and undefined at 𝑋, then 𝐵(𝑋) = 2o. (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝑋 ∈ On) ∧ ((𝐴 ↾ 𝑋) = (𝐵 ↾ 𝑋) ∧ 𝐴 <s 𝐵) ∧ (𝐴‘𝑋) = ∅) → (𝐵‘𝑋) = 2o)
Theorem	nogt01o 28083	Given 𝐴 greater than 𝐵, equal to 𝐵 up to 𝑋, and 𝐵(𝑋) undefined, then 𝐴(𝑋) = 1o. (Contributed by Scott Fenton, 9-Aug-2024.)
⊢ (((𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝑋 ∈ On) ∧ ((𝐴 ↾ 𝑋) = (𝐵 ↾ 𝑋) ∧ 𝐴 <s 𝐵) ∧ (𝐵‘𝑋) = ∅) → (𝐴‘𝑋) = 1o)
Theorem	noresle 28084*	Restriction law for surreals. Lemma 2.1.4 of [Lipparini] p. 3. (Contributed by Scott Fenton, 5-Dec-2021.)
⊢ (((𝑈 ∈ No ∧ 𝑆 ∈ No) ∧ (dom 𝑈 ⊆ 𝐴 ∧ dom 𝑆 ⊆ 𝐴 ∧ ∀𝑔 ∈ 𝐴 ¬ (𝑆 ↾ suc 𝑔) <s (𝑈 ↾ suc 𝑔))) → ¬ 𝑆 <s 𝑈)
Theorem	nomaxmo 28085*	A class of surreals has at most one maximum. (Contributed by Scott Fenton, 5-Dec-2021.)
⊢ (𝑆 ⊆ No → ∃*𝑥 ∈ 𝑆 ∀𝑦 ∈ 𝑆 ¬ 𝑥 <s 𝑦)
Theorem	nominmo 28086*	A class of surreals has at most one minimum. (Contributed by Scott Fenton, 8-Aug-2024.)
⊢ (𝑆 ⊆ No → ∃*𝑥 ∈ 𝑆 ∀𝑦 ∈ 𝑆 ¬ 𝑦 <s 𝑥)
Theorem	nosupprefixmo 28087*	In any class of surreals, there is at most one value of the prefix property. (Contributed by Scott Fenton, 26-Nov-2021.)
⊢ (𝐴 ⊆ No → ∃*𝑥∃𝑢 ∈ 𝐴 (𝐺 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝐺) = (𝑣 ↾ suc 𝐺)) ∧ (𝑢‘𝐺) = 𝑥))
Theorem	noinfprefixmo 28088*	In any class of surreals, there is at most one value of the prefix property. (Contributed by Scott Fenton, 8-Aug-2024.)
⊢ (𝐴 ⊆ No → ∃*𝑥∃𝑢 ∈ 𝐴 (𝐺 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑢 <s 𝑣 → (𝑢 ↾ suc 𝐺) = (𝑣 ↾ suc 𝐺)) ∧ (𝑢‘𝐺) = 𝑥))
Theorem	nosupcbv 28089*	Lemma to change bound variables in a surreal supremum. (Contributed by Scott Fenton, 9-Aug-2024.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ 𝑆 = if(∃𝑎 ∈ 𝐴 ∀𝑏 ∈ 𝐴 ¬ 𝑎 <s 𝑏, ((℩𝑎 ∈ 𝐴 ∀𝑏 ∈ 𝐴 ¬ 𝑎 <s 𝑏) ∪ {⟨dom (℩𝑎 ∈ 𝐴 ∀𝑏 ∈ 𝐴 ¬ 𝑎 <s 𝑏), 2o⟩}), (𝑐 ∈ {𝑑 ∣ ∃𝑒 ∈ 𝐴 (𝑑 ∈ dom 𝑒 ∧ ∀𝑓 ∈ 𝐴 (¬ 𝑓 <s 𝑒 → (𝑒 ↾ suc 𝑑) = (𝑓 ↾ suc 𝑑)))} ↦ (℩𝑎∃𝑒 ∈ 𝐴 (𝑐 ∈ dom 𝑒 ∧ ∀𝑓 ∈ 𝐴 (¬ 𝑓 <s 𝑒 → (𝑒 ↾ suc 𝑐) = (𝑓 ↾ suc 𝑐)) ∧ (𝑒‘𝑐) = 𝑎))))
Theorem	nosupno 28090*	The next several theorems deal with a surreal "supremum". This surreal will ultimately be shown to bound 𝐴 below and bound the restriction of any surreal above. We begin by showing that the given expression actually defines a surreal number. (Contributed by Scott Fenton, 5-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ ((𝐴 ⊆ No ∧ 𝐴 ∈ 𝑉) → 𝑆 ∈ No)
Theorem	nosupdm 28091*	The domain of the surreal supremum when there is no maximum. The primary point of this theorem is to change bound variable. (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ (¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦 → dom 𝑆 = {𝑧 ∣ ∃𝑝 ∈ 𝐴 (𝑧 ∈ dom 𝑝 ∧ ∀𝑞 ∈ 𝐴 (¬ 𝑞 <s 𝑝 → (𝑝 ↾ suc 𝑧) = (𝑞 ↾ suc 𝑧)))})
Theorem	nosupbday 28092*	Birthday bounding law for surreal supremum. (Contributed by Scott Fenton, 5-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ (((𝐴 ⊆ No ∧ 𝐴 ∈ V) ∧ (𝑂 ∈ On ∧ (bday “ 𝐴) ⊆ 𝑂)) → (bday‘𝑆) ⊆ 𝑂)
Theorem	nosupfv 28093*	The value of surreal supremum when there is no maximum. (Contributed by Scott Fenton, 5-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ ((¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦 ∧ (𝐴 ⊆ No ∧ 𝐴 ∈ V) ∧ (𝑈 ∈ 𝐴 ∧ 𝐺 ∈ dom 𝑈 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑈 → (𝑈 ↾ suc 𝐺) = (𝑣 ↾ suc 𝐺)))) → (𝑆‘𝐺) = (𝑈‘𝐺))
Theorem	nosupres 28094*	A restriction law for surreal supremum when there is no maximum. (Contributed by Scott Fenton, 5-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ ((¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦 ∧ (𝐴 ⊆ No ∧ 𝐴 ∈ V) ∧ (𝑈 ∈ 𝐴 ∧ 𝐺 ∈ dom 𝑈 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑈 → (𝑈 ↾ suc 𝐺) = (𝑣 ↾ suc 𝐺)))) → (𝑆 ↾ suc 𝐺) = (𝑈 ↾ suc 𝐺))
Theorem	nosupbnd1lem1 28095*	Lemma for nosupbnd1 28101. Establish a soft upper bound. (Contributed by Scott Fenton, 5-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ ((¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦 ∧ (𝐴 ⊆ No ∧ 𝐴 ∈ V) ∧ 𝑈 ∈ 𝐴) → ¬ 𝑆 <s (𝑈 ↾ dom 𝑆))
Theorem	nosupbnd1lem2 28096*	Lemma for nosupbnd1 28101. When there is no maximum, if any member of 𝐴 is a prolongment of 𝑆, then so are all elements of 𝐴 above it. (Contributed by Scott Fenton, 5-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ ((¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦 ∧ (𝐴 ⊆ No ∧ 𝐴 ∈ V) ∧ ((𝑈 ∈ 𝐴 ∧ (𝑈 ↾ dom 𝑆) = 𝑆) ∧ (𝑊 ∈ 𝐴 ∧ ¬ 𝑊 <s 𝑈))) → (𝑊 ↾ dom 𝑆) = 𝑆)
Theorem	nosupbnd1lem3 28097*	Lemma for nosupbnd1 28101. If 𝑈 is a prolongment of 𝑆 and in 𝐴, then (𝑈‘dom 𝑆) is not 2o. (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ ((¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦 ∧ (𝐴 ⊆ No ∧ 𝐴 ∈ V) ∧ (𝑈 ∈ 𝐴 ∧ (𝑈 ↾ dom 𝑆) = 𝑆)) → (𝑈‘dom 𝑆) ≠ 2o)
Theorem	nosupbnd1lem4 28098*	Lemma for nosupbnd1 28101. If 𝑈 is a prolongment of 𝑆 and in 𝐴, then (𝑈‘dom 𝑆) is not undefined. (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ ((¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦 ∧ (𝐴 ⊆ No ∧ 𝐴 ∈ V) ∧ (𝑈 ∈ 𝐴 ∧ (𝑈 ↾ dom 𝑆) = 𝑆)) → (𝑈‘dom 𝑆) ≠ ∅)
Theorem	nosupbnd1lem5 28099*	Lemma for nosupbnd1 28101. If 𝑈 is a prolongment of 𝑆 and in 𝐴, then (𝑈‘dom 𝑆) is not 1o. (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ ((¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦 ∧ (𝐴 ⊆ No ∧ 𝐴 ∈ V) ∧ (𝑈 ∈ 𝐴 ∧ (𝑈 ↾ dom 𝑆) = 𝑆)) → (𝑈‘dom 𝑆) ≠ 1o)
Theorem	nosupbnd1lem6 28100*	Lemma for nosupbnd1 28101. Establish a hard upper bound when there is no maximum. (Contributed by Scott Fenton, 6-Dec-2021.)
⊢ 𝑆 = if(∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦, ((℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦) ∪ {⟨dom (℩𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦), 2o⟩}), (𝑔 ∈ {𝑦 ∣ ∃𝑢 ∈ 𝐴 (𝑦 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑦) = (𝑣 ↾ suc 𝑦)))} ↦ (℩𝑥∃𝑢 ∈ 𝐴 (𝑔 ∈ dom 𝑢 ∧ ∀𝑣 ∈ 𝐴 (¬ 𝑣 <s 𝑢 → (𝑢 ↾ suc 𝑔) = (𝑣 ↾ suc 𝑔)) ∧ (𝑢‘𝑔) = 𝑥))))    ⇒   ⊢ ((¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐴 ¬ 𝑥 <s 𝑦 ∧ (𝐴 ⊆ No ∧ 𝐴 ∈ V) ∧ 𝑈 ∈ 𝐴) → (𝑈 ↾ dom 𝑆) <s 𝑆)