Theorem equsexvw 1919

Description: Version of equsexv 2095 with a dv condition, which requires fewer axioms. See also equsex 2281. (Contributed by BJ, 31-May-2019.)

Hypothesis

Ref	Expression
equsalvw.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

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

Distinct variable groups:   𝑥,𝑦   𝜓,𝑥

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

Proof of Theorem equsexvw

Step	Hyp	Ref	Expression
1		equsalvw.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	pm5.32i 667	. . 3 ⊢ ((𝑥 = 𝑦 ∧ 𝜑) ↔ (𝑥 = 𝑦 ∧ 𝜓))
3	2	exbii 1764	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ∃𝑥(𝑥 = 𝑦 ∧ 𝜓))
4		ax6ev 1877	. . 3 ⊢ ∃𝑥 𝑥 = 𝑦
5		19.41v 1901	. . 3 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜓) ↔ (∃𝑥 𝑥 = 𝑦 ∧ 𝜓))
6	4, 5	mpbiran 955	. 2 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜓) ↔ 𝜓)
7	3, 6	bitri 263	1 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 195   ∧ wa 383  ∃wex 1695

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875

This theorem depends on definitions:  df-bi 196  df-an 385  df-ex 1696

This theorem is referenced by:  cleljust  1985  sbhypf  3226  axsep  4708  dfid3  4954  opeliunxp  5093  imai  5397  coi1  5568  bj-axsep  31981