Theorem abbi 2134

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 2017	. 2 ⊢ ({x ∣ φ} = {x ∣ ψ} ↔ ∀y(y ∈ {x ∣ φ} ↔ y ∈ {x ∣ ψ}))
2		nfsab1 2013	. . . 4 ⊢ Ⅎx y ∈ {x ∣ φ}
3		nfsab1 2013	. . . 4 ⊢ Ⅎx y ∈ {x ∣ ψ}
4	2, 3	nfbi 1465	. . 3 ⊢ Ⅎx(y ∈ {x ∣ φ} ↔ y ∈ {x ∣ ψ})
5		nfv 1403	. . 3 ⊢ Ⅎy(φ ↔ ψ)
6		df-clab 2010	. . . . 5 ⊢ (y ∈ {x ∣ φ} ↔ [y / x]φ)
7		sbequ12r 1638	. . . . 5 ⊢ (y = x → ([y / x]φ ↔ φ))
8	6, 7	syl5bb 181	. . . 4 ⊢ (y = x → (y ∈ {x ∣ φ} ↔ φ))
9		df-clab 2010	. . . . 5 ⊢ (y ∈ {x ∣ ψ} ↔ [y / x]ψ)
10		sbequ12r 1638	. . . . 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 1620	. 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 1375  [wsb 1628  {cab 2009

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 1364  ax-ie2 1365  ax-8 1377  ax-11 1379  ax-4 1382  ax-17 1401  ax-i9 1405  ax-ial 1410  ax-i5r 1411  ax-ext 2005

This theorem depends on definitions:  df-bi 110  df-tru 1231  df-nf 1330  df-sb 1629  df-clab 2010  df-cleq 2016

This theorem is referenced by:  abbii  2136  abbid  2137  rabbi  2464  dfiota2  4793  iotabi  4801  uniabio  4802