Theorem equsalvw 2008

Description: Version of equsalv 2259 with a disjoint variable condition, and of equsal 2417 with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsexvw 2009. (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 1822	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
4		equsv 2007	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ 𝜓)
5	3, 4	bitri 275	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 205  ∀wal 1540

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1914  ax-6 1972

This theorem depends on definitions:  df-bi 206  df-ex 1783

This theorem is referenced by:  equsexvw  2009  equvelv  2035  sb6  2089  sbievw  2096  ax13lem2  2376  reu8  3730  el  5438  asymref2  6119  intirr  6120  fun11  6623  fv3  6910  fpwwe2lem11  10636  bj-dvelimdv  35730  bj-dvelimdv1  35731  undmrnresiss  42355  pm13.192  43169