Metamath Proof Explorer


Theorem eqfnfv2d2

Description: Equality of functions is determined by their values, a deduction version. (Contributed by metakunt, 28-May-2024)

Ref Expression
Hypotheses eqfnfv2d2.1 ⊢ ( 𝜑 → 𝐹 Fn 𝐴 )
eqfnfv2d2.2 ⊢ ( 𝜑 → 𝐺 Fn 𝐵 )
eqfnfv2d2.3 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
eqfnfv2d2.4 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) )
Assertion eqfnfv2d2 ( 𝜑 → 𝐹 = 𝐺 )

Proof

Step Hyp Ref Expression
1 eqfnfv2d2.1 ⊢ ( 𝜑 → 𝐹 Fn 𝐴 )
2 eqfnfv2d2.2 ⊢ ( 𝜑 → 𝐺 Fn 𝐵 )
3 eqfnfv2d2.3 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
4 eqfnfv2d2.4 ⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) )
5 4 ralrimiva ⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) )
6 3 5 jca ⊢ ( 𝜑 → ( 𝐴 = 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) )
7 1 2 jca ⊢ ( 𝜑 → ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵 ) )
8 eqfnfv2 ⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐵 ) → ( 𝐹 = 𝐺 ↔ ( 𝐴 = 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) ) )
9 7 8 syl ⊢ ( 𝜑 → ( 𝐹 = 𝐺 ↔ ( 𝐴 = 𝐵 ∧ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) ) )
10 6 9 mpbird ⊢ ( 𝜑 → 𝐹 = 𝐺 )