Theorem trcleq2lem 15028

Description: Equality implies bijection. (Contributed by RP, 5-May-2020.)

Assertion

Ref	Expression
trcleq2lem	⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Proof of Theorem trcleq2lem

Step	Hyp	Ref	Expression
1		sseq2 3971	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 ⊆ 𝐴 ↔ 𝑅 ⊆ 𝐵))
2		id 23	. . . 4 ⊢ (𝐴 = 𝐵 → 𝐴 = 𝐵)
3	2, 2	coeq12d 5851	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐴) = (𝐵 ∘ 𝐵))
4	3, 2	sseq12d 3978	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∘ 𝐴) ⊆ 𝐴 ↔ (𝐵 ∘ 𝐵) ⊆ 𝐵))
5	1, 4	anbi12d 643	1 ⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1567   ⊆ wss 3913   ∘ ccom 5666

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ss 3930  df-br 5114  df-opab 5178  df-co 5671

This theorem is referenced by:  cvbtrcl  15029  trcleq12lem  15030  trclublem  15032  cotrtrclfv  15049  trclun  15051  trclexi  44238  dftrcl3  44338