Theorem equsalvw 2006

Description: Version of equsalv 2264 with a disjoint variable condition, and of equsal 2435 with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsexvw 2007. (Contributed by BJ, 31-May-2019.)

Hypothesis

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

Assertion

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

Distinct variable groups:   𝑥,𝑦   𝜓,𝑥

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

Proof of Theorem equsalvw

Step	Hyp	Ref	Expression
1		equsalvw.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	pm5.74i 273	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜓))
3	2	albii 1816	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
4		equsv 2005	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ 𝜓)
5	3, 4	bitri 277	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 208  ∀wal 1531

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1907  ax-6 1966

This theorem depends on definitions:  df-bi 209  df-ex 1777

This theorem is referenced by:  equsexvw  2007  equvelv  2034  sb6  2089  sbievw  2099  ax13lem2  2390  reu8  3723  asymref2  5976  intirr  5977  fun11  6427  fv3  6687  fpwwe2lem12  10062  bj-dvelimdv  34175  bj-dvelimdv1  34176  wl-dfralflem  34837  undmrnresiss  39962  pm13.192  40740