Metamath Proof Explorer


Theorem kqfval

Description: Value of the function appearing in df-kq . (Contributed by Mario Carneiro, 25-Aug-2015)

Ref Expression
Hypothesis kqval.2 ⊢ 𝐹 = ( 𝑥 ∈ 𝑋 ↦ { 𝑦 ∈ 𝐽 ∣ 𝑥 ∈ 𝑦 } )
Assertion kqfval ( ( 𝐽 ∈ 𝑉 ∧ 𝐴 ∈ 𝑋 ) → ( 𝐹 ‘ 𝐴 ) = { 𝑦 ∈ 𝐽 ∣ 𝐴 ∈ 𝑦 } )

Proof

Step Hyp Ref Expression
1 kqval.2 ⊢ 𝐹 = ( 𝑥 ∈ 𝑋 ↦ { 𝑦 ∈ 𝐽 ∣ 𝑥 ∈ 𝑦 } )
2 id ⊢ ( 𝐴 ∈ 𝑋 → 𝐴 ∈ 𝑋 )
3 rabexg ⊢ ( 𝐽 ∈ 𝑉 → { 𝑦 ∈ 𝐽 ∣ 𝐴 ∈ 𝑦 } ∈ V )
4 eleq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 ∈ 𝑦 ↔ 𝐴 ∈ 𝑦 ) )
5 4 rabbidv ⊢ ( 𝑥 = 𝐴 → { 𝑦 ∈ 𝐽 ∣ 𝑥 ∈ 𝑦 } = { 𝑦 ∈ 𝐽 ∣ 𝐴 ∈ 𝑦 } )
6 5 1 fvmptg ⊢ ( ( 𝐴 ∈ 𝑋 ∧ { 𝑦 ∈ 𝐽 ∣ 𝐴 ∈ 𝑦 } ∈ V ) → ( 𝐹 ‘ 𝐴 ) = { 𝑦 ∈ 𝐽 ∣ 𝐴 ∈ 𝑦 } )
7 2 3 6 syl2anr ⊢ ( ( 𝐽 ∈ 𝑉 ∧ 𝐴 ∈ 𝑋 ) → ( 𝐹 ‘ 𝐴 ) = { 𝑦 ∈ 𝐽 ∣ 𝐴 ∈ 𝑦 } )