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Metamath Proof Explorer

Theorem feq2dd

Description: Equality deduction for functions. (Contributed by Thierry Arnoux, 27-May-2025)

Ref Expression
Hypotheses feq2dd.eq ⊢ ( 𝜑 → 𝐴 = 𝐵 )
feq2dd.f ⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝐶 )
Assertion feq2dd ( 𝜑 → 𝐹 : 𝐵 ⟶ 𝐶 )

Proof

Step Hyp Ref Expression
1 feq2dd.eq ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 feq2dd.f ⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝐶 )
3 1 feq2d ⊢ ( 𝜑 → ( 𝐹 : 𝐴 ⟶ 𝐶 ↔ 𝐹 : 𝐵 ⟶ 𝐶 ) )
4 2 3 mpbid ⊢ ( 𝜑 → 𝐹 : 𝐵 ⟶ 𝐶 )