Metamath Proof Explorer


Theorem currysetlem1

Description: Lemma for currysetALT . (Contributed by BJ, 23-Sep-2023) This proof is intuitionistically valid. (Proof modification is discouraged.)

Ref Expression
Hypothesis currysetlem2.def ⊢ 𝑋 = { 𝑥 ∣ ( 𝑥 ∈ 𝑥 → 𝜑 ) }
Assertion currysetlem1 ( 𝑋 ∈ 𝑉 → ( 𝑋 ∈ 𝑋 ↔ ( 𝑋 ∈ 𝑋 → 𝜑 ) ) )

Proof

Step Hyp Ref Expression
1 currysetlem2.def ⊢ 𝑋 = { 𝑥 ∣ ( 𝑥 ∈ 𝑥 → 𝜑 ) }
2 1 eqcomi ⊢ { 𝑥 ∣ ( 𝑥 ∈ 𝑥 → 𝜑 ) } = 𝑋
3 2 eleq2i ⊢ ( 𝑋 ∈ { 𝑥 ∣ ( 𝑥 ∈ 𝑥 → 𝜑 ) } ↔ 𝑋 ∈ 𝑋 )
4 nfab1 ⊢ Ⅎ 𝑥 { 𝑥 ∣ ( 𝑥 ∈ 𝑥 → 𝜑 ) }
5 1 4 nfcxfr ⊢ Ⅎ 𝑥 𝑋
6 5 5 nfel ⊢ Ⅎ 𝑥 𝑋 ∈ 𝑋
7 nfv ⊢ Ⅎ 𝑥 𝜑
8 6 7 nfim ⊢ Ⅎ 𝑥 ( 𝑋 ∈ 𝑋 → 𝜑 )
9 id ⊢ ( 𝑥 = 𝑋 → 𝑥 = 𝑋 )
10 9 9 eleq12d ⊢ ( 𝑥 = 𝑋 → ( 𝑥 ∈ 𝑥 ↔ 𝑋 ∈ 𝑋 ) )
11 10 imbi1d ⊢ ( 𝑥 = 𝑋 → ( ( 𝑥 ∈ 𝑥 → 𝜑 ) ↔ ( 𝑋 ∈ 𝑋 → 𝜑 ) ) )
12 5 8 11 elabgf ⊢ ( 𝑋 ∈ 𝑉 → ( 𝑋 ∈ { 𝑥 ∣ ( 𝑥 ∈ 𝑥 → 𝜑 ) } ↔ ( 𝑋 ∈ 𝑋 → 𝜑 ) ) )
13 3 12 bitr3id ⊢ ( 𝑋 ∈ 𝑉 → ( 𝑋 ∈ 𝑋 ↔ ( 𝑋 ∈ 𝑋 → 𝜑 ) ) )