Theorem scotteqd 9954

Description: Equality theorem for the Scott operation. (Contributed by Rohan Ridenour, 9-Aug-2023.)

Hypothesis

Ref	Expression
scotteqd.1	⊢ (𝜑 → 𝐴 = 𝐵)

Assertion

Ref	Expression
scotteqd	⊢ (𝜑 → Scott 𝐴 = Scott 𝐵)

Proof of Theorem scotteqd

Dummy variables 𝑥 𝑦 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		scotteqd.1	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
2	1	adantr 486	. . . 4 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → 𝐴 = 𝐵)
3	2	raleqdv 3323	. . 3 ⊢ ((𝜑 ∧ 𝑥 ∈ 𝐴) → (∀𝑦 ∈ 𝐴 (rank‘𝑥) ⊆ (rank‘𝑦) ↔ ∀𝑦 ∈ 𝐵 (rank‘𝑥) ⊆ (rank‘𝑦)))
4	1, 3	rabeqbidva 3432	. 2 ⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ ∀𝑦 ∈ 𝐴 (rank‘𝑥) ⊆ (rank‘𝑦)} = {𝑥 ∈ 𝐵 ∣ ∀𝑦 ∈ 𝐵 (rank‘𝑥) ⊆ (rank‘𝑦)})
5		df-scott 9953	. 2 ⊢ Scott 𝐴 = {𝑥 ∈ 𝐴 ∣ ∀𝑦 ∈ 𝐴 (rank‘𝑥) ⊆ (rank‘𝑦)}
6		df-scott 9953	. 2 ⊢ Scott 𝐵 = {𝑥 ∈ 𝐵 ∣ ∀𝑦 ∈ 𝐵 (rank‘𝑥) ⊆ (rank‘𝑦)}
7	4, 5, 6	3eqtr4g 2823	1 ⊢ (𝜑 → Scott 𝐴 = Scott 𝐵)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2147  ∀wral 3080  {crab 3416   ⊆ wss 3900  ‘cfv 6555  rankcrnk 9791  Scott cscott 9952

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-scott 9953

This theorem is used by:  scotteq  9955  karden  9983  erscottd  35842  ecscottd  35843  kardval  35919  kard0  35921  dfcoll2  45426  colleq12d  45427