Theorem equsalvw 2005

Description: Version of equsalv 2270 with a disjoint variable condition, and of equsal 2417 with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsexvw 2006. (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 1820	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
4		equsv 2004	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ 𝜓)
5	3, 4	bitri 275	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1539

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968

This theorem depends on definitions:  df-bi 207  df-ex 1781

This theorem is referenced by:  equsexvw  2006  equvelv  2032  sb6  2088  sbievwOLD  2097  ax13lem2  2376  reu8  3687  el  5378  asymref2  6063  intirr  6064  fun11  6555  fv3  6840  elirrv  9483  fpwwe2lem11  10532  axreg  35125  axregscl  35126  bj-dvelimdv  36895  bj-dvelimdv1  36896  undmrnresiss  43696  pm13.192  44502