Theorem equsexvw 2038

Description: Version of equsexv 2304 with a disjoint variable condition, and of equsex 2449 with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsalvw 2037. (Contributed by BJ, 31-May-2019.) (Proof shortened by Wolf Lammen, 23-Oct-2023.)

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		alinexa 1876	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → ¬ 𝜑) ↔ ¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑))
2		equsalvw.1	. . . . 5 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	2	notbid 321	. . . 4 ⊢ (𝑥 = 𝑦 → (¬ 𝜑 ↔ ¬ 𝜓))
4	3	equsalvw 2037	. . 3 ⊢ (∀𝑥(𝑥 = 𝑦 → ¬ 𝜑) ↔ ¬ 𝜓)
5	1, 4	bitr3i 280	. 2 ⊢ (¬ ∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ ¬ 𝜓)
6	5	con4bii 324	1 ⊢ (∃𝑥(𝑥 = 𝑦 ∧ 𝜑) ↔ 𝜓)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812

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 1943  ax-6 2000

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813

This theorem is used by:  equvinv  2062  cleljust  2154  sbelx  2290  cleljustab  2743  axsepgfromrep  5253  dfid3  5557  opeliunxp  5726  opeliun2xp  5727  imai  6074  coi1  6263  opabex3d  7965  opabex3rd  7966  opabex3  7967  fsplit  8117  mapsnend  9046  elirrv  9572  elirrvOLD  9573  dfac5lem1  10129  dfac5lem3  10131  dffix2  36469  sscoid  36477  elfuns  36479  pmapglb  40630  polval2N  40766  tfsconcat0i  44173