Theorem elabf1 14930

Description: One implication of elabf 2895. (Contributed by BJ, 21-Nov-2019.)

Hypotheses

Ref	Expression
elabf1.nf	⊢ Ⅎ𝑥𝜓
elabf1.1	⊢ (𝑥 = 𝐴 → (𝜑 → 𝜓))

Assertion

Ref	Expression
elabf1	⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} → 𝜓)

Distinct variable group:   𝑥,𝐴

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

Proof of Theorem elabf1

Step	Hyp	Ref	Expression
1		nfcv 2332	. 2 ⊢ Ⅎ𝑥𝐴
2		elabf1.nf	. 2 ⊢ Ⅎ𝑥𝜓
3		elabf1.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 → 𝜓))
4	1, 2, 3	elabgf1 14928	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} → 𝜓)

Syntax hints:   → wi 4   = wceq 1364  Ⅎwnf 1471   ∈ wcel 2160  {cab 2175

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2171

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2176  df-cleq 2182  df-clel 2185  df-nfc 2321  df-v 2754

This theorem is referenced by:  elab1  14932  bj-bdfindis  15096