Theorem abeq1 2188

Description: Equality of a class variable and a class abstraction. (Contributed by NM, 20-Aug-1993.)

Assertion

Ref	Expression
abeq1	⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ ∀𝑥(𝜑 ↔ 𝑥 ∈ 𝐴))

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem abeq1

Step	Hyp	Ref	Expression
1		abeq2 2187	. 2 ⊢ (𝐴 = {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑))
2		eqcom 2083	. 2 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ 𝐴 = {𝑥 ∣ 𝜑})
3		bicom 138	. . 3 ⊢ ((𝜑 ↔ 𝑥 ∈ 𝐴) ↔ (𝑥 ∈ 𝐴 ↔ 𝜑))
4	3	albii 1399	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝑥 ∈ 𝐴) ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝜑))
5	1, 2, 4	3bitr4i 210	1 ⊢ ({𝑥 ∣ 𝜑} = 𝐴 ↔ ∀𝑥(𝜑 ↔ 𝑥 ∈ 𝐴))

Syntax hints:   ↔ wb 103  ∀wal 1282   = wceq 1284   ∈ wcel 1433  {cab 2067

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-11 1437  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077

This theorem is referenced by:  abbi1dv  2198  disj  3292  euabsn2  3461  dm0rn0  4570  dffo3  5335