Theorem trcleq2lem 15040

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 4035	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 ⊆ 𝐴 ↔ 𝑅 ⊆ 𝐵))
2		id 22	. . . 4 ⊢ (𝐴 = 𝐵 → 𝐴 = 𝐵)
3	2, 2	coeq12d 5889	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐴) = (𝐵 ∘ 𝐵))
4	3, 2	sseq12d 4042	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∘ 𝐴) ⊆ 𝐴 ↔ (𝐵 ∘ 𝐵) ⊆ 𝐵))
5	1, 4	anbi12d 631	1 ⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1537   ⊆ wss 3976   ∘ ccom 5704

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2711

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1778  df-sb 2065  df-clab 2718  df-cleq 2732  df-clel 2819  df-ss 3993  df-br 5167  df-opab 5229  df-co 5709

This theorem is referenced by:  cvbtrcl  15041  trcleq12lem  15042  trclublem  15044  cotrtrclfv  15061  trclun  15063  trclexi  43582  dftrcl3  43682