Theorem elscottrankeq 35676

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 488	. . 3 ⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → 𝐴 ∈ Scott 𝐶)
2		simpr 490	. . 3 ⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → 𝐵 ∈ Scott 𝐶)
3	1, 2	scottelrankd 9919	. 2 ⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → (rank‘𝐴) ⊆ (rank‘𝐵))
4	2, 1	scottelrankd 9919	. 2 ⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → (rank‘𝐵) ⊆ (rank‘𝐴))
5	3, 4	eqssd 3947	1 ⊢ ((𝐴 ∈ Scott 𝐶 ∧ 𝐵 ∈ Scott 𝐶) → (rank‘𝐴) = (rank‘𝐵))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145  ‘cfv 6527  rankcrnk 9745  Scott cscott 9899

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-ral 3077  df-rab 3413  df-v 3452  df-dif 3901  df-un 3903  df-ss 3915  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-uni 4867  df-br 5103  df-iota 6483  df-fv 6535  df-scott 9900

This theorem is used by: (None)