Theorem trcleq12lem 13941

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 13930	. 2 ⊢ (𝑅 = 𝑆 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴)))
2		trcleq2lem 13939	. 2 ⊢ (𝐴 = 𝐵 → ((𝑆 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))
3	1, 2	sylan9bb 493	1 ⊢ ((𝑅 = 𝑆 ∧ 𝐴 = 𝐵) → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑆 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 382   = wceq 1630   ⊆ wss 3721   ∘ ccom 5253

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1869  ax-4 1884  ax-5 1990  ax-6 2056  ax-7 2092  ax-9 2153  ax-10 2173  ax-11 2189  ax-12 2202  ax-13 2407  ax-ext 2750

This theorem depends on definitions:  df-bi 197  df-an 383  df-or 827  df-tru 1633  df-ex 1852  df-nf 1857  df-sb 2049  df-clab 2757  df-cleq 2763  df-clel 2766  df-nfc 2901  df-in 3728  df-ss 3735  df-br 4785  df-opab 4845  df-co 5258

This theorem is referenced by: (None)