Theorem elabgf1 9918

Description: One implication of elabgf 2685. (Contributed by BJ, 21-Nov-2019.)

Hypotheses

Ref	Expression
elabgf1.nf1	⊢ Ⅎ𝑥𝐴
elabgf1.nf2	⊢ Ⅎ𝑥𝜓
elabgf1.1	⊢ (𝑥 = 𝐴 → (𝜑 → 𝜓))

Assertion

Ref	Expression
elabgf1	⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} → 𝜓)

Proof of Theorem elabgf1

Step	Hyp	Ref	Expression
1		elabgf1.nf1	. . 3 ⊢ Ⅎ𝑥𝐴
2		elabgf1.nf2	. . 3 ⊢ Ⅎ𝑥𝜓
3	1, 2	elabgft1 9917	. 2 ⊢ (∀𝑥(𝑥 = 𝐴 → (𝜑 → 𝜓)) → (𝐴 ∈ {𝑥 ∣ 𝜑} → 𝜓))
4		elabgf1.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 → 𝜓))
5	3, 4	mpg 1340	1 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} → 𝜓)

Syntax hints:   → wi 4   = wceq 1243  Ⅎwnf 1349   ∈ wcel 1393  {cab 2026  Ⅎwnfc 2165

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559

This theorem is referenced by:  elabf1  9920