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Metamath Proof Explorer

Theorem suppvalfng

Description: The value of the operation constructing the support of a function with a given domain. This version of suppvalfn assumes F is a set rather than its domain X , avoiding ax-rep . (Contributed by SN, 5-Aug-2024)

Ref Expression
Assertion suppvalfng ( ( 𝐹 Fn 𝑋 ∧ 𝐹 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → ( 𝐹 supp 𝑍 ) = { 𝑖 ∈ 𝑋 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } )

Proof

Step Hyp Ref Expression
1 fnfun ⊢ ( 𝐹 Fn 𝑋 → Fun 𝐹 )
2 suppval1 ⊢ ( ( Fun 𝐹 ∧ 𝐹 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → ( 𝐹 supp 𝑍 ) = { 𝑖 ∈ dom 𝐹 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } )
3 1 2 syl3an1 ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝐹 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → ( 𝐹 supp 𝑍 ) = { 𝑖 ∈ dom 𝐹 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } )
4 fndm ⊢ ( 𝐹 Fn 𝑋 → dom 𝐹 = 𝑋 )
5 4 3ad2ant1 ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝐹 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → dom 𝐹 = 𝑋 )
6 5 rabeqdv ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝐹 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → { 𝑖 ∈ dom 𝐹 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } = { 𝑖 ∈ 𝑋 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } )
7 3 6 eqtrd ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝐹 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → ( 𝐹 supp 𝑍 ) = { 𝑖 ∈ 𝑋 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } )