Theorem elabf1 12988

Description: One implication of elabf 2827. (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 2281	. 2 ⊢ Ⅎ𝑥𝐴
2		elabf1.nf	. 2 ⊢ Ⅎ𝑥𝜓
3		elabf1.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 → 𝜓))
4	1, 2, 3	elabgf1 12986	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} → 𝜓)

Syntax hints:   → wi 4   = wceq 1331  Ⅎwnf 1436   ∈ wcel 1480  {cab 2125

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688

This theorem is referenced by:  elab1  12990  bj-bdfindis  13145