Metamath Proof Explorer


Theorem f13idfv

Description: A one-to-one function with the domain { 0, 1 ,2 } in terms of function values. (Contributed by Alexander van der Vekens, 26-Jan-2018)

Ref Expression
Hypothesis f13idfv.a 𝐴 = ( 0 ... 2 )
Assertion f13idfv ( 𝐹 : 𝐴1-1𝐵 ↔ ( 𝐹 : 𝐴𝐵 ∧ ( ( 𝐹 ‘ 0 ) ≠ ( 𝐹 ‘ 1 ) ∧ ( 𝐹 ‘ 0 ) ≠ ( 𝐹 ‘ 2 ) ∧ ( 𝐹 ‘ 1 ) ≠ ( 𝐹 ‘ 2 ) ) ) )

Proof

Step Hyp Ref Expression
1 f13idfv.a 𝐴 = ( 0 ... 2 )
2 0z 0 ∈ ℤ
3 1z 1 ∈ ℤ
4 2z 2 ∈ ℤ
5 2 3 4 3pm3.2i ( 0 ∈ ℤ ∧ 1 ∈ ℤ ∧ 2 ∈ ℤ )
6 0ne1 0 ≠ 1
7 0ne2 0 ≠ 2
8 1ne2 1 ≠ 2
9 6 7 8 3pm3.2i ( 0 ≠ 1 ∧ 0 ≠ 2 ∧ 1 ≠ 2 )
10 fz0tp ( 0 ... 2 ) = { 0 , 1 , 2 }
11 1 10 eqtri 𝐴 = { 0 , 1 , 2 }
12 11 f13dfv ( ( ( 0 ∈ ℤ ∧ 1 ∈ ℤ ∧ 2 ∈ ℤ ) ∧ ( 0 ≠ 1 ∧ 0 ≠ 2 ∧ 1 ≠ 2 ) ) → ( 𝐹 : 𝐴1-1𝐵 ↔ ( 𝐹 : 𝐴𝐵 ∧ ( ( 𝐹 ‘ 0 ) ≠ ( 𝐹 ‘ 1 ) ∧ ( 𝐹 ‘ 0 ) ≠ ( 𝐹 ‘ 2 ) ∧ ( 𝐹 ‘ 1 ) ≠ ( 𝐹 ‘ 2 ) ) ) ) )
13 5 9 12 mp2an ( 𝐹 : 𝐴1-1𝐵 ↔ ( 𝐹 : 𝐴𝐵 ∧ ( ( 𝐹 ‘ 0 ) ≠ ( 𝐹 ‘ 1 ) ∧ ( 𝐹 ‘ 0 ) ≠ ( 𝐹 ‘ 2 ) ∧ ( 𝐹 ‘ 1 ) ≠ ( 𝐹 ‘ 2 ) ) ) )