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Metamath Proof Explorer

Theorem fndmdifeq0

Description: The difference set of two functions is empty if and only if the functions are equal. (Contributed by Stefan O'Rear, 17-Jan-2015)

Ref Expression
Assertion fndmdifeq0 ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( dom ( 𝐹 ∖ 𝐺 ) = ∅ ↔ 𝐹 = 𝐺 ) )

Proof

Step Hyp Ref Expression
1 fndmdif ⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → dom ( 𝐹 ∖ 𝐺 ) = { 𝑥 ∈ 𝐴 ∣ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) } )
2 1 eqeq1d ⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( dom ( 𝐹 ∖ 𝐺 ) = ∅ ↔ { 𝑥 ∈ 𝐴 ∣ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) } = ∅ ) )
3 rabeq0 ⊢ ( { 𝑥 ∈ 𝐴 ∣ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) } = ∅ ↔ ∀ 𝑥 ∈ 𝐴 ¬ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) )
4 nne ⊢ ( ¬ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) ↔ ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) )
5 4 ralbii ⊢ ( ∀ 𝑥 ∈ 𝐴 ¬ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) )
6 3 5 bitri ⊢ ( { 𝑥 ∈ 𝐴 ∣ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) } = ∅ ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) )
7 eqfnfv ⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( 𝐹 = 𝐺 ↔ ∀ 𝑥 ∈ 𝐴 ( 𝐹 ‘ 𝑥 ) = ( 𝐺 ‘ 𝑥 ) ) )
8 6 7 bitr4id ⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( { 𝑥 ∈ 𝐴 ∣ ( 𝐹 ‘ 𝑥 ) ≠ ( 𝐺 ‘ 𝑥 ) } = ∅ ↔ 𝐹 = 𝐺 ) )
9 2 8 bitrd ⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴 ) → ( dom ( 𝐹 ∖ 𝐺 ) = ∅ ↔ 𝐹 = 𝐺 ) )