Theorem elscottrankeq 35524

Description: Elements in a Scott's trick set have the same rank. (Contributed by BTernaryTau, 9-Jul-2026.)

Assertion

Ref	Expression
elscottrankeq	⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → (rank‘𝐴) = (rank‘𝐵))

Proof of Theorem elscottrankeq

Step	Hyp	Ref	Expression
1		simpl 487	. . 3 ⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → 𝐴 ∈ Scott 𝐶)
2		simpr 489	. . 3 ⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → 𝐵 ∈ Scott 𝐶)
3	1, 2	scottelrankd 9875	. 2 ⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → (rank‘𝐴) ⊆ (rank‘𝐵))
4	2, 1	scottelrankd 9875	. 2 ⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → (rank‘𝐵) ⊆ (rank‘𝐴))
5	3, 4	eqssd 3953	1 ⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → (rank‘𝐴) = (rank‘𝐵))

This proof depends on syntax axioms:   → wi 4   ∧ wa 400   = wceq 1569   ∈ wcel 2142  ‘cfv 6536  rankcrnk 9733  Scott cscott 9855

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-fal 1582  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-rab 3416  df-v 3456  df-dif 3907  df-un 3909  df-ss 3921  df-nul 4286  df-if 4487  df-sn 4589  df-pr 4591  df-op 4595  df-uni 4872  df-br 5109  df-iota 6492  df-fv 6544  df-scott 9856

This theorem is used by: (None)