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Metamath Proof Explorer

Theorem f1veqaeq

Description: If the values of a one-to-one function for two arguments are equal, the arguments themselves must be equal. (Contributed by Alexander van der Vekens, 12-Nov-2017)

Ref Expression
Assertion f1veqaeq ( ( 𝐹 : 𝐴 –1-1→ 𝐵 ∧ ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐴 ) ) → ( ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝐷 ) → 𝐶 = 𝐷 ) )

Proof

Step Hyp Ref Expression
1 dff13 ⊢ ( 𝐹 : 𝐴 –1-1→ 𝐵 ↔ ( 𝐹 : 𝐴 ⟶ 𝐵 ∧ ∀ 𝑐 ∈ 𝐴 ∀ 𝑑 ∈ 𝐴 ( ( 𝐹 ‘ 𝑐 ) = ( 𝐹 ‘ 𝑑 ) → 𝑐 = 𝑑 ) ) )
2 fveqeq2 ⊢ ( 𝑐 = 𝐶 → ( ( 𝐹 ‘ 𝑐 ) = ( 𝐹 ‘ 𝑑 ) ↔ ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝑑 ) ) )
3 eqeq1 ⊢ ( 𝑐 = 𝐶 → ( 𝑐 = 𝑑 ↔ 𝐶 = 𝑑 ) )
4 2 3 imbi12d ⊢ ( 𝑐 = 𝐶 → ( ( ( 𝐹 ‘ 𝑐 ) = ( 𝐹 ‘ 𝑑 ) → 𝑐 = 𝑑 ) ↔ ( ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝑑 ) → 𝐶 = 𝑑 ) ) )
5 fveq2 ⊢ ( 𝑑 = 𝐷 → ( 𝐹 ‘ 𝑑 ) = ( 𝐹 ‘ 𝐷 ) )
6 5 eqeq2d ⊢ ( 𝑑 = 𝐷 → ( ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝑑 ) ↔ ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝐷 ) ) )
7 eqeq2 ⊢ ( 𝑑 = 𝐷 → ( 𝐶 = 𝑑 ↔ 𝐶 = 𝐷 ) )
8 6 7 imbi12d ⊢ ( 𝑑 = 𝐷 → ( ( ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝑑 ) → 𝐶 = 𝑑 ) ↔ ( ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝐷 ) → 𝐶 = 𝐷 ) ) )
9 4 8 rspc2v ⊢ ( ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐴 ) → ( ∀ 𝑐 ∈ 𝐴 ∀ 𝑑 ∈ 𝐴 ( ( 𝐹 ‘ 𝑐 ) = ( 𝐹 ‘ 𝑑 ) → 𝑐 = 𝑑 ) → ( ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝐷 ) → 𝐶 = 𝐷 ) ) )
10 9 com12 ⊢ ( ∀ 𝑐 ∈ 𝐴 ∀ 𝑑 ∈ 𝐴 ( ( 𝐹 ‘ 𝑐 ) = ( 𝐹 ‘ 𝑑 ) → 𝑐 = 𝑑 ) → ( ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐴 ) → ( ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝐷 ) → 𝐶 = 𝐷 ) ) )
11 1 10 simplbiim ⊢ ( 𝐹 : 𝐴 –1-1→ 𝐵 → ( ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐴 ) → ( ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝐷 ) → 𝐶 = 𝐷 ) ) )
12 11 imp ⊢ ( ( 𝐹 : 𝐴 –1-1→ 𝐵 ∧ ( 𝐶 ∈ 𝐴 ∧ 𝐷 ∈ 𝐴 ) ) → ( ( 𝐹 ‘ 𝐶 ) = ( 𝐹 ‘ 𝐷 ) → 𝐶 = 𝐷 ) )