Metamath Proof Explorer


Theorem fvdifsupp

Description: Function value is zero outside of its support. (Contributed by Thierry Arnoux, 21-Jan-2024)

Ref Expression
Hypotheses fvdifsupp.1 ( 𝜑𝐹 Fn 𝐴 )
fvdifsupp.2 ( 𝜑𝐴𝑉 )
fvdifsupp.3 ( 𝜑𝑍𝑊 )
fvdifsupp.4 ( 𝜑𝑋 ∈ ( 𝐴 ∖ ( 𝐹 supp 𝑍 ) ) )
Assertion fvdifsupp ( 𝜑 → ( 𝐹𝑋 ) = 𝑍 )

Proof

Step Hyp Ref Expression
1 fvdifsupp.1 ( 𝜑𝐹 Fn 𝐴 )
2 fvdifsupp.2 ( 𝜑𝐴𝑉 )
3 fvdifsupp.3 ( 𝜑𝑍𝑊 )
4 fvdifsupp.4 ( 𝜑𝑋 ∈ ( 𝐴 ∖ ( 𝐹 supp 𝑍 ) ) )
5 4 eldifbd ( 𝜑 → ¬ 𝑋 ∈ ( 𝐹 supp 𝑍 ) )
6 4 eldifad ( 𝜑𝑋𝐴 )
7 elsuppfn ( ( 𝐹 Fn 𝐴𝐴𝑉𝑍𝑊 ) → ( 𝑋 ∈ ( 𝐹 supp 𝑍 ) ↔ ( 𝑋𝐴 ∧ ( 𝐹𝑋 ) ≠ 𝑍 ) ) )
8 1 2 3 7 syl3anc ( 𝜑 → ( 𝑋 ∈ ( 𝐹 supp 𝑍 ) ↔ ( 𝑋𝐴 ∧ ( 𝐹𝑋 ) ≠ 𝑍 ) ) )
9 6 8 mpbirand ( 𝜑 → ( 𝑋 ∈ ( 𝐹 supp 𝑍 ) ↔ ( 𝐹𝑋 ) ≠ 𝑍 ) )
10 9 necon2bbid ( 𝜑 → ( ( 𝐹𝑋 ) = 𝑍 ↔ ¬ 𝑋 ∈ ( 𝐹 supp 𝑍 ) ) )
11 5 10 mpbird ( 𝜑 → ( 𝐹𝑋 ) = 𝑍 )