Theorem abbibOLD 2833

Description: Obsolete version of abbib 2832 as of 22-Oct-2026. (Contributed by NM, 25-Nov-2013.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

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

Proof of Theorem abbibOLD

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfcleq 2756	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}))
2		nfsab1 2749	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ {𝑥 ∣ 𝜑}
3		nfsab1 2749	. . . 4 ⊢ Ⅎ𝑥 𝑦 ∈ {𝑥 ∣ 𝜓}
4	2, 3	nfbi 1938	. . 3 ⊢ Ⅎ𝑥(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓})
5		nfv 1949	. . 3 ⊢ Ⅎ𝑦(𝜑 ↔ 𝜓)
6		df-clab 2742	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
7		sbequ12r 2291	. . . . 5 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜑 ↔ 𝜑))
8	6, 7	bitrid 286	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜑))
9		df-clab 2742	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ [𝑦 / 𝑥]𝜓)
10		sbequ12r 2291	. . . . 5 ⊢ (𝑦 = 𝑥 → ([𝑦 / 𝑥]𝜓 ↔ 𝜓))
11	9, 10	bitrid 286	. . . 4 ⊢ (𝑦 = 𝑥 → (𝑦 ∈ {𝑥 ∣ 𝜓} ↔ 𝜓))
12	8, 11	bibi12d 348	. . 3 ⊢ (𝑦 = 𝑥 → ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ (𝜑 ↔ 𝜓)))
13	4, 5, 12	cbvalv1 2374	. 2 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ 𝜓}) ↔ ∀𝑥(𝜑 ↔ 𝜓))
14	1, 13	bitri 278	1 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} ↔ ∀𝑥(𝜑 ↔ 𝜓))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1568   = wceq 1570  [wsb 2101   ∈ wcel 2147  {cab 2741

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1945  ax-6 2002  ax-7 2043  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2102  df-clab 2742  df-cleq 2755

This theorem is used by: (None)