Metamath Proof Explorer


Theorem fnreseql

Description: Two functions are equal on a subset iff their equalizer contains that subset. (Contributed by Stefan O'Rear, 7-Mar-2015)

Ref Expression
Assertion fnreseql ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴𝑋𝐴 ) → ( ( 𝐹𝑋 ) = ( 𝐺𝑋 ) ↔ 𝑋 ⊆ dom ( 𝐹𝐺 ) ) )

Proof

Step Hyp Ref Expression
1 fnssres ( ( 𝐹 Fn 𝐴𝑋𝐴 ) → ( 𝐹𝑋 ) Fn 𝑋 )
2 1 3adant2 ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴𝑋𝐴 ) → ( 𝐹𝑋 ) Fn 𝑋 )
3 fnssres ( ( 𝐺 Fn 𝐴𝑋𝐴 ) → ( 𝐺𝑋 ) Fn 𝑋 )
4 3 3adant1 ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴𝑋𝐴 ) → ( 𝐺𝑋 ) Fn 𝑋 )
5 fneqeql ( ( ( 𝐹𝑋 ) Fn 𝑋 ∧ ( 𝐺𝑋 ) Fn 𝑋 ) → ( ( 𝐹𝑋 ) = ( 𝐺𝑋 ) ↔ dom ( ( 𝐹𝑋 ) ∩ ( 𝐺𝑋 ) ) = 𝑋 ) )
6 2 4 5 syl2anc ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴𝑋𝐴 ) → ( ( 𝐹𝑋 ) = ( 𝐺𝑋 ) ↔ dom ( ( 𝐹𝑋 ) ∩ ( 𝐺𝑋 ) ) = 𝑋 ) )
7 resindir ( ( 𝐹𝐺 ) ↾ 𝑋 ) = ( ( 𝐹𝑋 ) ∩ ( 𝐺𝑋 ) )
8 7 dmeqi dom ( ( 𝐹𝐺 ) ↾ 𝑋 ) = dom ( ( 𝐹𝑋 ) ∩ ( 𝐺𝑋 ) )
9 dmres dom ( ( 𝐹𝐺 ) ↾ 𝑋 ) = ( 𝑋 ∩ dom ( 𝐹𝐺 ) )
10 8 9 eqtr3i dom ( ( 𝐹𝑋 ) ∩ ( 𝐺𝑋 ) ) = ( 𝑋 ∩ dom ( 𝐹𝐺 ) )
11 10 eqeq1i ( dom ( ( 𝐹𝑋 ) ∩ ( 𝐺𝑋 ) ) = 𝑋 ↔ ( 𝑋 ∩ dom ( 𝐹𝐺 ) ) = 𝑋 )
12 df-ss ( 𝑋 ⊆ dom ( 𝐹𝐺 ) ↔ ( 𝑋 ∩ dom ( 𝐹𝐺 ) ) = 𝑋 )
13 11 12 bitr4i ( dom ( ( 𝐹𝑋 ) ∩ ( 𝐺𝑋 ) ) = 𝑋𝑋 ⊆ dom ( 𝐹𝐺 ) )
14 6 13 bitrdi ( ( 𝐹 Fn 𝐴𝐺 Fn 𝐴𝑋𝐴 ) → ( ( 𝐹𝑋 ) = ( 𝐺𝑋 ) ↔ 𝑋 ⊆ dom ( 𝐹𝐺 ) ) )