Theorem elabg 2972

Description: Membership in a class abstraction, using implicit substitution. Compare Theorem 6.13 of [Quine] p. 44. (Contributed by NM, 14-Apr-1995.)

Hypothesis

Ref	Expression
elabg.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
elabg	⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

Distinct variable groups:   𝜓,𝑥   𝑥,𝐴

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

Proof of Theorem elabg

Step	Hyp	Ref	Expression
1		nfcv 2392	. 2 ⊢ Ⅎ𝑥𝐴
2		nfv 1581	. 2 ⊢ Ⅎ𝑥𝜓
3		elabg.1	. 2 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
4	1, 2, 3	elabgf 2968	1 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 105   = wceq 1402   ∈ wcel 2209  {cab 2224

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823

This theorem is used by:  elab2g  2973  intmin3  3997  finds  4747  elxpi  4790  elabrexg  5964  ovelrn  6238  elabreximd  6356  elfi  7305  indpi  7709  peano5nnnn  8259  peano5nni  9308  hashf1lem1  11287  lss1d  14722  lspsn  14755  zndvds  14986  eltg  15155  eltg2  15156  ausgrusgrien  16424