Metamath Proof Explorer


Theorem trcleq2lemRP

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

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

Proof

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