Metamath Proof Explorer


Theorem fvpr2g

Description: The value of a function with a domain of (at most) two elements. (Contributed by Alexander van der Vekens, 3-Dec-2017) (Proof shortened by BJ, 26-Sep-2024)

Ref Expression
Assertion fvpr2g ( ( 𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵 ) → ( { ⟨ 𝐴 , 𝐶 ⟩ , ⟨ 𝐵 , 𝐷 ⟩ } ‘ 𝐵 ) = 𝐷 )

Proof

Step Hyp Ref Expression
1 prcom ⊢ { ⟨ 𝐴 , 𝐶 ⟩ , ⟨ 𝐵 , 𝐷 ⟩ } = { ⟨ 𝐵 , 𝐷 ⟩ , ⟨ 𝐴 , 𝐶 ⟩ }
2 1 fveq1i ⊢ ( { ⟨ 𝐴 , 𝐶 ⟩ , ⟨ 𝐵 , 𝐷 ⟩ } ‘ 𝐵 ) = ( { ⟨ 𝐵 , 𝐷 ⟩ , ⟨ 𝐴 , 𝐶 ⟩ } ‘ 𝐵 )
3 necom ⊢ ( 𝐴 ≠ 𝐵 ↔ 𝐵 ≠ 𝐴 )
4 fvpr1g ⊢ ( ( 𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ∧ 𝐵 ≠ 𝐴 ) → ( { ⟨ 𝐵 , 𝐷 ⟩ , ⟨ 𝐴 , 𝐶 ⟩ } ‘ 𝐵 ) = 𝐷 )
5 3 4 syl3an3b ⊢ ( ( 𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵 ) → ( { ⟨ 𝐵 , 𝐷 ⟩ , ⟨ 𝐴 , 𝐶 ⟩ } ‘ 𝐵 ) = 𝐷 )
6 2 5 eqtrid ⊢ ( ( 𝐵 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵 ) → ( { ⟨ 𝐴 , 𝐶 ⟩ , ⟨ 𝐵 , 𝐷 ⟩ } ‘ 𝐵 ) = 𝐷 )