Theorem elab3g 3642

Description: Membership in a class abstraction, with a weaker antecedent than elabg 3633. (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 3633	. . . 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 3633	. 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 1570   ∈ wcel 2145  {cab 2740

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837

This theorem is used by:  elab3  3643  elssabg  5311  elrnmptg  5949  elrelimasn  6086  elmapg  8842  isust  24436  ellimc  26107  isismty  38559  clublem  44458