Theorem trcleq12lem 15158

Description: Equality implies bijection. (Contributed by RP, 9-May-2020.)

Assertion

Ref	Expression
trcleq12lem	⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Proof of Theorem trcleq12lem

Step	Hyp	Ref	Expression
1		cleq1lem 15147	. 2 ⊢ (𝑅 = 𝑆 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴)))
2		trcleq2lem 15156	. 2 ⊢ (𝐴 = 𝐵 → ((𝑆 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))
3	1, 2	sylan9bb 519	1 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1569   ⊆ wss 3899   ∘ ccom 5658

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-8 2148  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-clel 2838  df-ss 3900  df-br 5106  df-opab 5170  df-co 5663

This theorem is used by: (None)