Theorem abbi 2133

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 2016	. 2 ⊢ ({x ∣ φ} = {x ∣ ψ} ↔ ∀y(y ∈ {x ∣ φ} ↔ y ∈ {x ∣ ψ}))
2		nfsab1 2012	. . . 4 ⊢ Ⅎx y ∈ {x ∣ φ}
3		nfsab1 2012	. . . 4 ⊢ Ⅎx y ∈ {x ∣ ψ}
4	2, 3	nfbi 1463	. . 3 ⊢ Ⅎx(y ∈ {x ∣ φ} ↔ y ∈ {x ∣ ψ})
5		nfv 1402	. . 3 ⊢ Ⅎy(φ ↔ ψ)
6		df-clab 2009	. . . . 5 ⊢ (y ∈ {x ∣ φ} ↔ [y / x]φ)
7		sbequ12r 1637	. . . . 5 ⊢ (y = x → ([y / x]φ ↔ φ))
8	6, 7	syl5bb 181	. . . 4 ⊢ (y = x → (y ∈ {x ∣ φ} ↔ φ))
9		df-clab 2009	. . . . 5 ⊢ (y ∈ {x ∣ ψ} ↔ [y / x]ψ)
10		sbequ12r 1637	. . . . 5 ⊢ (y = x → ([y / x]ψ ↔ ψ))
11	9, 10	syl5bb 181	. . . 4 ⊢ (y = x → (y ∈ {x ∣ ψ} ↔ ψ))
12	8, 11	bibi12d 224	. . 3 ⊢ (y = x → ((y ∈ {x ∣ φ} ↔ y ∈ {x ∣ ψ}) ↔ (φ ↔ ψ)))
13	4, 5, 12	cbval 1619	. 2 ⊢ (∀y(y ∈ {x ∣ φ} ↔ y ∈ {x ∣ ψ}) ↔ ∀x(φ ↔ ψ))
14	1, 13	bitr2i 174	1 ⊢ (∀x(φ ↔ ψ) ↔ {x ∣ φ} = {x ∣ ψ})

Syntax hints:   ↔ wb 98  ∀wal 1226   = wceq 1228   ∈ wcel 1374  [wsb 1627  {cab 2008

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-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-11 1378  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004

This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015

This theorem is referenced by:  abbii  2135  abbid  2136  rabbi  2463  dfiota2  4793  iotabi  4801  uniabio  4802