Theorem abbi 2319

Description: Equivalent wff's correspond to equal class abstractions. (Contributed by NM, 25-Nov-2013.) (Revised by Mario Carneiro, 11-Aug-2016.)

Assertion

Ref	Expression
abbi	⊢ (∀𝑥(𝜑 ↔ 𝜓) ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓})

Proof of Theorem abbi

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfcleq 2199	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}))
2		nfsab1 2195	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ {𝑥 ∣ 𝜑}
3		nfsab1 2195	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ {𝑥 ∣ 𝜓}
4	2, 3	nfbi 1612	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓})
5		nfv 1551	. . 3 ⊢ Ⅎ𝑦(𝜑 ↔ 𝜓)
6		df-clab 2192	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
7		sbequ12r 1795	. . . . 5 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑))
8	6, 7	bitrid 192	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))
9		df-clab 2192	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
10		sbequ12r 1795	. . . . 5 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜓 ↔ 𝜓))
11	9, 10	bitrid 192	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜓))
12	8, 11	bibi12d 235	. . 3 ⊢ (𝑦 = 𝑥 → ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜑 ↔ 𝜓)))
13	4, 5, 12	cbval 1777	. 2 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑥(𝜑 ↔ 𝜓))
14	1, 13	bitr2i 185	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓})

Syntax hints:   ↔ wb 105  ∀wal 1371   = wceq 1373  [wsb 1785   ∈ wcel 2176  {cab 2191

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-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-11 1529  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-ext 2187

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1484  df-sb 1786  df-clab 2192  df-cleq 2198

This theorem is referenced by:  abbii  2321  abbid  2322  rabbi  2684  sbcbi2  3049  dfiota2  5233  iotabi  5241  uniabio  5242