Theorem equsalvw 2004

Description: Version of equsalv 2268 with a disjoint variable condition, and of equsal 2415 with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsexvw 2005. (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 271	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜓))
3	2	albii 1819	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
4		equsv 2003	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ 𝜓)
5	3, 4	bitri 275	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1538

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967

This theorem depends on definitions:  df-bi 207  df-ex 1780

This theorem is referenced by:  equsexvw  2005  equvelv  2031  sb6  2086  sbievwOLD  2095  ax13lem2  2374  reu8  3693  el  5381  asymref2  6066  intirr  6067  fun11  6556  fv3  6840  elirrv  9489  fpwwe2lem11  10535  axreg  35068  axregscl  35069  bj-dvelimdv  36835  bj-dvelimdv1  36836  undmrnresiss  43587  pm13.192  44393