Theorem equsexv 2291

Description: Version of equsex 2425 with a disjoint variable condition, which does not require ax-13 2377. See equsexvw 2104 for a version with two disjoint variable conditions requiring fewer axioms. See also the dual form equsalv 2290. (Contributed 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 571	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) ↔ (𝑥 = 𝑦 ∧ 𝜓))
3	2	exbii 1944	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓))
4		ax6ev 2074	. . 3 ⊢ ∃𝑥 𝑥 = 𝑦
5		equsalv.nf	. . . 4 ⊢ Ⅎ𝑥𝜓
6	5	19.41 2270	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜓) ↔ (∃𝑥 𝑥 = 𝑦 ∧ 𝜓))
7	4, 6	mpbiran 701	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜓) ↔ 𝜓)
8	3, 7	bitri 267	1 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 385  ∃wex 1875  Ⅎwnf 1879

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-12 2213

This theorem depends on definitions:  df-bi 199  df-an 386  df-ex 1876  df-nf 1880

This theorem is referenced by:  sb56  2297  equsexhv  2316  cleljustALT2  2373  sb10f  2576  dprd2d2  18759  poimirlem25  33923