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Metamath Proof Explorer

Theorem suppssrg

Description: A function is zero outside its support. Version of suppssr avoiding ax-rep by assuming F is a set rather than its domain A . (Contributed by SN, 5-May-2024)

Ref Expression
Hypotheses suppssrg.f ⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝐵 )
suppssrg.n ⊢ ( 𝜑 → ( 𝐹 supp 𝑍 ) ⊆ 𝑊 )
suppssrg.a ⊢ ( 𝜑 → 𝐹 ∈ 𝑉 )
suppssrg.z ⊢ ( 𝜑 → 𝑍 ∈ 𝑈 )
Assertion suppssrg ( ( 𝜑 ∧ 𝑋 ∈ ( 𝐴 ∖ 𝑊 ) ) → ( 𝐹 ‘ 𝑋 ) = 𝑍 )

Proof

Step Hyp Ref Expression
1 suppssrg.f ⊢ ( 𝜑 → 𝐹 : 𝐴 ⟶ 𝐵 )
2 suppssrg.n ⊢ ( 𝜑 → ( 𝐹 supp 𝑍 ) ⊆ 𝑊 )
3 suppssrg.a ⊢ ( 𝜑 → 𝐹 ∈ 𝑉 )
4 suppssrg.z ⊢ ( 𝜑 → 𝑍 ∈ 𝑈 )
5 eldif ⊢ ( 𝑋 ∈ ( 𝐴 ∖ 𝑊 ) ↔ ( 𝑋 ∈ 𝐴 ∧ ¬ 𝑋 ∈ 𝑊 ) )
6 1 ffnd ⊢ ( 𝜑 → 𝐹 Fn 𝐴 )
7 elsuppfng ⊢ ( ( 𝐹 Fn 𝐴 ∧ 𝐹 ∈ 𝑉 ∧ 𝑍 ∈ 𝑈 ) → ( 𝑋 ∈ ( 𝐹 supp 𝑍 ) ↔ ( 𝑋 ∈ 𝐴 ∧ ( 𝐹 ‘ 𝑋 ) ≠ 𝑍 ) ) )
8 6 3 4 7 syl3anc ⊢ ( 𝜑 → ( 𝑋 ∈ ( 𝐹 supp 𝑍 ) ↔ ( 𝑋 ∈ 𝐴 ∧ ( 𝐹 ‘ 𝑋 ) ≠ 𝑍 ) ) )
9 2 sseld ⊢ ( 𝜑 → ( 𝑋 ∈ ( 𝐹 supp 𝑍 ) → 𝑋 ∈ 𝑊 ) )
10 8 9 sylbird ⊢ ( 𝜑 → ( ( 𝑋 ∈ 𝐴 ∧ ( 𝐹 ‘ 𝑋 ) ≠ 𝑍 ) → 𝑋 ∈ 𝑊 ) )
11 10 expdimp ⊢ ( ( 𝜑 ∧ 𝑋 ∈ 𝐴 ) → ( ( 𝐹 ‘ 𝑋 ) ≠ 𝑍 → 𝑋 ∈ 𝑊 ) )
12 11 necon1bd ⊢ ( ( 𝜑 ∧ 𝑋 ∈ 𝐴 ) → ( ¬ 𝑋 ∈ 𝑊 → ( 𝐹 ‘ 𝑋 ) = 𝑍 ) )
13 12 impr ⊢ ( ( 𝜑 ∧ ( 𝑋 ∈ 𝐴 ∧ ¬ 𝑋 ∈ 𝑊 ) ) → ( 𝐹 ‘ 𝑋 ) = 𝑍 )
14 5 13 sylan2b ⊢ ( ( 𝜑 ∧ 𝑋 ∈ ( 𝐴 ∖ 𝑊 ) ) → ( 𝐹 ‘ 𝑋 ) = 𝑍 )