Metamath Proof Explorer


Theorem fvtp2

Description: The second value of a function with a domain of three elements. (Contributed by NM, 14-Sep-2011)

Ref Expression
Hypotheses fvtp2.1 ⊢ 𝐵 ∈ V
fvtp2.4 ⊢ 𝐸 ∈ V
Assertion fvtp2 ( ( 𝐴 ≠ 𝐵 ∧ 𝐵 ≠ 𝐶 ) → ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ } ‘ 𝐵 ) = 𝐸 )

Proof

Step Hyp Ref Expression
1 fvtp2.1 ⊢ 𝐵 ∈ V
2 fvtp2.4 ⊢ 𝐸 ∈ V
3 tprot ⊢ { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ } = { ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ , ⟨ 𝐴 , 𝐷 ⟩ }
4 3 fveq1i ⊢ ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ } ‘ 𝐵 ) = ( { ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ , ⟨ 𝐴 , 𝐷 ⟩ } ‘ 𝐵 )
5 necom ⊢ ( 𝐴 ≠ 𝐵 ↔ 𝐵 ≠ 𝐴 )
6 1 2 fvtp1 ⊢ ( ( 𝐵 ≠ 𝐶 ∧ 𝐵 ≠ 𝐴 ) → ( { ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ , ⟨ 𝐴 , 𝐷 ⟩ } ‘ 𝐵 ) = 𝐸 )
7 6 ancoms ⊢ ( ( 𝐵 ≠ 𝐴 ∧ 𝐵 ≠ 𝐶 ) → ( { ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ , ⟨ 𝐴 , 𝐷 ⟩ } ‘ 𝐵 ) = 𝐸 )
8 5 7 sylanb ⊢ ( ( 𝐴 ≠ 𝐵 ∧ 𝐵 ≠ 𝐶 ) → ( { ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ , ⟨ 𝐴 , 𝐷 ⟩ } ‘ 𝐵 ) = 𝐸 )
9 4 8 eqtrid ⊢ ( ( 𝐴 ≠ 𝐵 ∧ 𝐵 ≠ 𝐶 ) → ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ } ‘ 𝐵 ) = 𝐸 )