Theorem elab3g 3638

Description: Membership in a class abstraction, with a weaker antecedent than elabg 3629. (Contributed by NM, 29-Aug-2006.)

Hypothesis

Ref	Expression
elab3g.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
elab3g	⊢ ((𝜓 → 𝐴 ∈ 𝐵) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Distinct variable groups:   𝜓,𝑥   𝑥,𝐴

Allowed substitution hints:   𝜑(𝑥)   𝐵(𝑥)

Proof of Theorem elab3g

Step	Hyp	Ref	Expression
1		elab3g.1	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
2	1	elabg 3629	. . . 4 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))
3	2	ibi 270	. . 3 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} → 𝜓)
4		pm2.21 124	. . 3 ⊢ (¬ 𝜓 → (𝜓 → 𝐴 ∈ {𝑥 ∣ 𝜑}))
5	3, 4	impbid2 229	. 2 ⊢ (¬ 𝜓 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))
6	1	elabg 3629	. 2 ⊢ (𝐴 ∈ 𝐵 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))
7	5, 6	ja 188	1 ⊢ ((𝜓 → 𝐴 ∈ 𝐵) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   = wceq 1569   ∈ wcel 2144  {cab 2738

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835

This theorem is used by:  elab3  3639  elssabg  5303  elrnmptg  5942  elrelimasn  6083  elmapg  8863  isust  24529  ellimc  26199  isismty  38829  clublem  44704