Theorem trcleq2lemRP 42314

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

Assertion

Ref	Expression
trcleq2lemRP	⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Proof of Theorem trcleq2lemRP

Step	Hyp	Ref	Expression
1		id 22	. . . 4 ⊢ (𝐴 = 𝐵 → 𝐴 = 𝐵)
2	1, 1	coeq12d 5862	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐴) = (𝐵 ∘ 𝐵))
3	2, 1	sseq12d 4014	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∘ 𝐴) ⊆ 𝐴 ↔ (𝐵 ∘ 𝐵) ⊆ 𝐵))
4	3	cleq2lem 42292	1 ⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 397   = wceq 1542   ⊆ wss 3947   ∘ ccom 5679

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2109  ax-9 2117  ax-ext 2704

This theorem depends on definitions:  df-bi 206  df-an 398  df-tru 1545  df-ex 1783  df-sb 2069  df-clab 2711  df-cleq 2725  df-clel 2811  df-v 3477  df-in 3954  df-ss 3964  df-br 5148  df-opab 5210  df-co 5684

This theorem is referenced by: (None)