Theorem trcleq2lem 15027

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 4022	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 ⊆ 𝐴 ↔ 𝑅 ⊆ 𝐵))
2		id 22	. . . 4 ⊢ (𝐴 = 𝐵 → 𝐴 = 𝐵)
3	2, 2	coeq12d 5878	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐴) = (𝐵 ∘ 𝐵))
4	3, 2	sseq12d 4029	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∘ 𝐴) ⊆ 𝐴 ↔ (𝐵 ∘ 𝐵) ⊆ 𝐵))
5	1, 4	anbi12d 632	1 ⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1537   ⊆ wss 3963   ∘ ccom 5693

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1777  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-ss 3980  df-br 5149  df-opab 5211  df-co 5698

This theorem is referenced by:  cvbtrcl  15028  trcleq12lem  15029  trclublem  15031  cotrtrclfv  15048  trclun  15050  trclexi  43610  dftrcl3  43710