Theorem equsalvw 2003

Description: Version of equsalv 2268 with a disjoint variable condition, and of equsal 2425 with two disjoint variable conditions, which requires fewer axioms. See also the dual form equsexvw 2004. (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 1817	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
4		equsv 2002	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ 𝜓)
5	3, 4	bitri 275	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1535

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967

This theorem depends on definitions:  df-bi 207  df-ex 1778

This theorem is referenced by:  equsexvw  2004  equvelv  2030  sb6  2085  sbievwOLD  2094  ax13lem2  2384  reu8  3755  el  5457  asymref2  6149  intirr  6150  fun11  6652  fv3  6938  fpwwe2lem11  10710  bj-dvelimdv  36817  bj-dvelimdv1  36818  undmrnresiss  43566  pm13.192  44379