Theorem elabf1 14415

Description: One implication of elabf 2880. (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 2319	. 2 ⊢ Ⅎ𝑥𝐴
2		elabf1.nf	. 2 ⊢ Ⅎ𝑥𝜓
3		elabf1.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 → 𝜓))
4	1, 2, 3	elabgf1 14413	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} → 𝜓)

Syntax hints:   → wi 4   = wceq 1353  Ⅎwnf 1460   ∈ wcel 2148  {cab 2163

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2739

This theorem is referenced by:  elab1  14417  bj-bdfindis  14581