Theorem trcleq2lem 15054

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 3964	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 ⊆ 𝐴 ↔ 𝑅 ⊆ 𝐵))
2		id 23	. . . 4 ⊢ (𝐴 = 𝐵 → 𝐴 = 𝐵)
3	2, 2	coeq12d 5852	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐴) = (𝐵 ∘ 𝐵))
4	3, 2	sseq12d 3971	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∘ 𝐴) ⊆ 𝐴 ↔ (𝐵 ∘ 𝐵) ⊆ 𝐵))
5	1, 4	anbi12d 644	1 ⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ⊆ wss 3906   ∘ ccom 5667

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 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-ss 3923  df-br 5112  df-opab 5176  df-co 5672

This theorem is used by:  cvbtrcl  15055  trcleq12lem  15056  trclublem  15058  cotrtrclfv  15075  trclun  15077  trclexi  44406  dftrcl3  44506