Theorem abbi 2280

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 2159	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}))
2		nfsab1 2155	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ {𝑥 ∣ 𝜑}
3		nfsab1 2155	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ {𝑥 ∣ 𝜓}
4	2, 3	nfbi 1577	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓})
5		nfv 1516	. . 3 ⊢ Ⅎ𝑦(𝜑 ↔ 𝜓)
6		df-clab 2152	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
7		sbequ12r 1760	. . . . 5 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑))
8	6, 7	syl5bb 191	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))
9		df-clab 2152	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
10		sbequ12r 1760	. . . . 5 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜓 ↔ 𝜓))
11	9, 10	syl5bb 191	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜓))
12	8, 11	bibi12d 234	. . 3 ⊢ (𝑦 = 𝑥 → ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜑 ↔ 𝜓)))
13	4, 5, 12	cbval 1742	. 2 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑥(𝜑 ↔ 𝜓))
14	1, 13	bitr2i 184	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓})

Syntax hints:   ↔ wb 104  ∀wal 1341   = wceq 1343  [wsb 1750   ∈ wcel 2136  {cab 2151

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-11 1494  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158

This theorem is referenced by:  abbii  2282  abbid  2283  rabbi  2643  sbcbi2  3001  dfiota2  5154  iotabi  5162  uniabio  5163