Metamath Proof Explorer


Theorem fdifsuppconst

Description: A function is a zero constant outside of its support. (Contributed by Thierry Arnoux, 22-Jun-2024)

Ref Expression
Hypothesis fdifsuppconst.1 𝐴 = ( dom 𝐹 ∖ ( 𝐹 supp 𝑍 ) )
Assertion fdifsuppconst ( ( Fun 𝐹𝐹𝑉𝑍𝑊 ) → ( 𝐹𝐴 ) = ( 𝐴 × { 𝑍 } ) )

Proof

Step Hyp Ref Expression
1 fdifsuppconst.1 𝐴 = ( dom 𝐹 ∖ ( 𝐹 supp 𝑍 ) )
2 funfn ( Fun 𝐹𝐹 Fn dom 𝐹 )
3 2 biimpi ( Fun 𝐹𝐹 Fn dom 𝐹 )
4 3 ad2antrr ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) → 𝐹 Fn dom 𝐹 )
5 difssd ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) → ( dom 𝐹 ∖ ( 𝐹 supp 𝑍 ) ) ⊆ dom 𝐹 )
6 1 5 eqsstrid ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) → 𝐴 ⊆ dom 𝐹 )
7 4 6 fnssresd ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) → ( 𝐹𝐴 ) Fn 𝐴 )
8 fnconstg ( 𝑍𝑊 → ( 𝐴 × { 𝑍 } ) Fn 𝐴 )
9 8 adantl ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) → ( 𝐴 × { 𝑍 } ) Fn 𝐴 )
10 4 adantr ( ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) ∧ 𝑥𝐴 ) → 𝐹 Fn dom 𝐹 )
11 dmexg ( 𝐹𝑉 → dom 𝐹 ∈ V )
12 11 ad3antlr ( ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) ∧ 𝑥𝐴 ) → dom 𝐹 ∈ V )
13 simplr ( ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) ∧ 𝑥𝐴 ) → 𝑍𝑊 )
14 1 eleq2i ( 𝑥𝐴𝑥 ∈ ( dom 𝐹 ∖ ( 𝐹 supp 𝑍 ) ) )
15 14 biimpi ( 𝑥𝐴𝑥 ∈ ( dom 𝐹 ∖ ( 𝐹 supp 𝑍 ) ) )
16 15 adantl ( ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) ∧ 𝑥𝐴 ) → 𝑥 ∈ ( dom 𝐹 ∖ ( 𝐹 supp 𝑍 ) ) )
17 10 12 13 16 fvdifsupp ( ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) ∧ 𝑥𝐴 ) → ( 𝐹𝑥 ) = 𝑍 )
18 simpr ( ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) ∧ 𝑥𝐴 ) → 𝑥𝐴 )
19 18 fvresd ( ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) ∧ 𝑥𝐴 ) → ( ( 𝐹𝐴 ) ‘ 𝑥 ) = ( 𝐹𝑥 ) )
20 fvconst2g ( ( 𝑍𝑊𝑥𝐴 ) → ( ( 𝐴 × { 𝑍 } ) ‘ 𝑥 ) = 𝑍 )
21 20 adantll ( ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) ∧ 𝑥𝐴 ) → ( ( 𝐴 × { 𝑍 } ) ‘ 𝑥 ) = 𝑍 )
22 17 19 21 3eqtr4d ( ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) ∧ 𝑥𝐴 ) → ( ( 𝐹𝐴 ) ‘ 𝑥 ) = ( ( 𝐴 × { 𝑍 } ) ‘ 𝑥 ) )
23 7 9 22 eqfnfvd ( ( ( Fun 𝐹𝐹𝑉 ) ∧ 𝑍𝑊 ) → ( 𝐹𝐴 ) = ( 𝐴 × { 𝑍 } ) )
24 23 3impa ( ( Fun 𝐹𝐹𝑉𝑍𝑊 ) → ( 𝐹𝐴 ) = ( 𝐴 × { 𝑍 } ) )