Theorem trcleq12lem 14944

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 14933	. 2 ⊢ (𝑅 = 𝑆 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴)))
2		trcleq2lem 14942	. 2 ⊢ (𝐴 = 𝐵 → ((𝑆 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))
3	1, 2	sylan9bb 508	1 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 394   = wceq 1539   ⊆ wss 3947   ∘ ccom 5679

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1911  ax-6 1969  ax-7 2009  ax-8 2106  ax-9 2114  ax-ext 2701

This theorem depends on definitions:  df-bi 206  df-an 395  df-tru 1542  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2722  df-clel 2808  df-v 3474  df-in 3954  df-ss 3964  df-br 5148  df-opab 5210  df-co 5684

This theorem is referenced by: (None)