Theorem cleqabd 1178

Description: Equality of a class variable and a class abstraction (deduction).

Hypothesis

Ref	Expression
cleqabd.1	⊢ (φ → A = {x∣ψ})

Assertion

Ref	Expression
cleqabd	⊢ (φ → (x ∈ A ↔ ψ))

Proof of Theorem cleqabd

Step	Hyp	Ref	Expression
1		cleqabd.1	. . 3 ⊢ (φ → A = {x∣ψ})
2	1	eleq2d 1156	. 2 ⊢ (φ → (x ∈ A ↔ x ∈ {x∣ψ}))
3		abid 1094	. 2 ⊢ (x ∈ {x∣ψ} ↔ ψ)
4	2, 3	syl6bb 414	1 ⊢ (φ → (x ∈ A ↔ ψ))

Syntax hints:   → wi 2   ↔ wb 127  {cab 1090   = wceq 1091   ∈ wcel 1092

This theorem is referenced by:  genpn0 3900  genpss 3901  genpnmax 3904

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-gen 677  ax-9 799  ax-12 802  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099

metamath.org