Metamath Proof Explorer


Theorem trcleq2lem

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

Ref Expression
Assertion trcleq2lem ( 𝐴 = 𝐵 → ( ( 𝑅 ⊆ 𝐴 ∧ ( 𝐴 ∘ 𝐴 ) ⊆ 𝐴 ) ↔ ( 𝑅 ⊆ 𝐵 ∧ ( 𝐵 ∘ 𝐵 ) ⊆ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 sseq2 ⊢ ( 𝐴 = 𝐵 → ( 𝑅 ⊆ 𝐴 ↔ 𝑅 ⊆ 𝐵 ) )
2 id ⊢ ( 𝐴 = 𝐵 → 𝐴 = 𝐵 )
3 2 2 coeq12d ⊢ ( 𝐴 = 𝐵 → ( 𝐴 ∘ 𝐴 ) = ( 𝐵 ∘ 𝐵 ) )
4 3 2 sseq12d ⊢ ( 𝐴 = 𝐵 → ( ( 𝐴 ∘ 𝐴 ) ⊆ 𝐴 ↔ ( 𝐵 ∘ 𝐵 ) ⊆ 𝐵 ) )
5 1 4 anbi12d ⊢ ( 𝐴 = 𝐵 → ( ( 𝑅 ⊆ 𝐴 ∧ ( 𝐴 ∘ 𝐴 ) ⊆ 𝐴 ) ↔ ( 𝑅 ⊆ 𝐵 ∧ ( 𝐵 ∘ 𝐵 ) ⊆ 𝐵 ) ) )