Theorem trcleq12lem 15026

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 15015	. 2 ⊢ (𝑅 = 𝑆 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴)))
2		trcleq2lem 15024	. 2 ⊢ (𝐴 = 𝐵 → ((𝑆 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))
3	1, 2	sylan9bb 518	1 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   = wceq 1570   ⊆ wss 3905   ∘ ccom 5665

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-ss 3922  df-br 5110  df-opab 5174  df-co 5670

This theorem is referenced by: (None)