Theorem trcleq2lemRP 44159

Description: Equality implies bijection. (Contributed by RP, 5-May-2020.) (Proof modification is discouraged.)

Assertion

Ref	Expression
trcleq2lemRP	⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Proof of Theorem trcleq2lemRP

Step	Hyp	Ref	Expression
1		id 22	. . . 4 ⊢ (𝐴 = 𝐵 → 𝐴 = 𝐵)
2	1, 1	coeq12d 5834	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∘ 𝐴) = (𝐵 ∘ 𝐵))
3	2, 1	sseq12d 3969	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∘ 𝐴) ⊆ 𝐴 ↔ (𝐵 ∘ 𝐵) ⊆ 𝐵))
4	3	cleq2lem 44137	1 ⊢ (𝐴 = 𝐵 → ((𝑅 ⊆ 𝐴 ∧ (𝐴 ∘ 𝐴) ⊆ 𝐴) ↔ (𝑅 ⊆ 𝐵 ∧ (𝐵 ∘ 𝐵) ⊆ 𝐵)))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399   = wceq 1559   ⊆ wss 3904   ∘ ccom 5649

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1799  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-ss 3921  df-br 5100  df-opab 5162  df-co 5654

This theorem is referenced by: (None)