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Metamath Proof Explorer

Theorem fvtp1g

Description: The value of a function with a domain of (at most) three elements. (Contributed by Alexander van der Vekens, 4-Dec-2017)

Ref Expression
Assertion fvtp1g ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ) ∧ ( 𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ) ) → ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ } ‘ 𝐴 ) = 𝐷 )

Proof

Step Hyp Ref Expression
1 df-tp ⊢ { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ } = ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ∪ { ⟨ 𝐶 , 𝐹 ⟩ } )
2 1 fveq1i ⊢ ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ } ‘ 𝐴 ) = ( ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ∪ { ⟨ 𝐶 , 𝐹 ⟩ } ) ‘ 𝐴 )
3 necom ⊢ ( 𝐴 ≠ 𝐶 ↔ 𝐶 ≠ 𝐴 )
4 fvunsn ⊢ ( 𝐶 ≠ 𝐴 → ( ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ∪ { ⟨ 𝐶 , 𝐹 ⟩ } ) ‘ 𝐴 ) = ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ‘ 𝐴 ) )
5 3 4 sylbi ⊢ ( 𝐴 ≠ 𝐶 → ( ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ∪ { ⟨ 𝐶 , 𝐹 ⟩ } ) ‘ 𝐴 ) = ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ‘ 𝐴 ) )
6 5 ad2antll ⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ) ∧ ( 𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ) ) → ( ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ∪ { ⟨ 𝐶 , 𝐹 ⟩ } ) ‘ 𝐴 ) = ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ‘ 𝐴 ) )
7 fvpr1g ⊢ ( ( 𝐴 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ∧ 𝐴 ≠ 𝐵 ) → ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ‘ 𝐴 ) = 𝐷 )
8 7 3expa ⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ) ∧ 𝐴 ≠ 𝐵 ) → ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ‘ 𝐴 ) = 𝐷 )
9 8 adantrr ⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ) ∧ ( 𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ) ) → ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ‘ 𝐴 ) = 𝐷 )
10 6 9 eqtrd ⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ) ∧ ( 𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ) ) → ( ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ } ∪ { ⟨ 𝐶 , 𝐹 ⟩ } ) ‘ 𝐴 ) = 𝐷 )
11 2 10 eqtrid ⊢ ( ( ( 𝐴 ∈ 𝑉 ∧ 𝐷 ∈ 𝑊 ) ∧ ( 𝐴 ≠ 𝐵 ∧ 𝐴 ≠ 𝐶 ) ) → ( { ⟨ 𝐴 , 𝐷 ⟩ , ⟨ 𝐵 , 𝐸 ⟩ , ⟨ 𝐶 , 𝐹 ⟩ } ‘ 𝐴 ) = 𝐷 )