Theorem trcleq2lem 15187

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 3939	. 2 ⊢ (𝐴 = 𝐵 → (𝑅 ⊆ 𝐴 ↔ 𝑅 ⊆ 𝐵))
2		id 23	. . . 4 ⊢ (𝐴 = 𝐵 → 𝐴 = 𝐵)
3	2, 2	coeq12d 5853	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐴) = (𝐵 ∘ 𝐵))
4	3, 2	sseq12d 3946	. 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 3900   ∘ ccom 5662

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ss 3901  df-br 5109  df-opab 5174  df-co 5667

This theorem is used by:  cvbtrcl  15188  trcleq12lem  15189  trclublem  15191  cotrtrclfv  15208  trclun  15210  trclexi  44881  dftrcl3  44981