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Metamath Proof Explorer

Theorem suppvalfn

Description: The value of the operation constructing the support of a function with a given domain. (Contributed by Stefan O'Rear, 1-Feb-2015) (Revised by AV, 22-Apr-2019)

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

Proof

Step Hyp Ref Expression
1 fnfun ⊢ ( 𝐹 Fn 𝑋 → Fun 𝐹 )
2 1 3ad2ant1 ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝑋 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → Fun 𝐹 )
3 fnex ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝑋 ∈ 𝑉 ) → 𝐹 ∈ V )
4 3 3adant3 ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝑋 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → 𝐹 ∈ V )
5 simp3 ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝑋 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → 𝑍 ∈ 𝑊 )
6 suppval1 ⊢ ( ( Fun 𝐹 ∧ 𝐹 ∈ V ∧ 𝑍 ∈ 𝑊 ) → ( 𝐹 supp 𝑍 ) = { 𝑖 ∈ dom 𝐹 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } )
7 2 4 5 6 syl3anc ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝑋 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → ( 𝐹 supp 𝑍 ) = { 𝑖 ∈ dom 𝐹 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } )
8 fndm ⊢ ( 𝐹 Fn 𝑋 → dom 𝐹 = 𝑋 )
9 8 3ad2ant1 ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝑋 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → dom 𝐹 = 𝑋 )
10 9 rabeqdv ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝑋 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → { 𝑖 ∈ dom 𝐹 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } = { 𝑖 ∈ 𝑋 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } )
11 7 10 eqtrd ⊢ ( ( 𝐹 Fn 𝑋 ∧ 𝑋 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊 ) → ( 𝐹 supp 𝑍 ) = { 𝑖 ∈ 𝑋 ∣ ( 𝐹 ‘ 𝑖 ) ≠ 𝑍 } )