Theorem equsexv 2269

Description: An equivalence related to implicit substitution. Version of equsex 2440 with a disjoint variable condition, which does not require ax-13 2390. See equsexvw 2011 for a version with two disjoint variable conditions requiring fewer axioms. See also the dual form equsalv 2268. (Contributed by NM, 5-Aug-1993.) (Revised by BJ, 31-May-2019.)

Hypotheses

Ref	Expression
equsalv.nf	⊢ Ⅎ𝑥𝜓
equsalv.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
equsexv	⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ 𝜓)

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑥,𝑦)

Proof of Theorem equsexv

Step	Hyp	Ref	Expression
1		equsalv.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	pm5.32i 577	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) ↔ (𝑥 = 𝑦 ∧ 𝜓))
3	2	exbii 1848	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓))
4		ax6ev 1972	. . 3 ⊢ ∃𝑥 𝑥 = 𝑦
5		equsalv.nf	. . . 4 ⊢ Ⅎ𝑥𝜓
6	5	19.41 2237	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜓) ↔ (∃𝑥 𝑥 = 𝑦 ∧ 𝜓))
7	4, 6	mpbiran 707	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜓) ↔ 𝜓)
8	3, 7	bitri 277	1 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398  ∃wex 1780  Ⅎwnf 1784

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-12 2177

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1781  df-nf 1785

This theorem is referenced by:  sb5  2276  sb56OLD  2278  equsexhv  2300  cleljustALT2  2383  sb10f  2571  dprd2d2  19168  poimirlem25  34919