Theorem equsalvw 2006

Description: Version of equsalv 2275 with a disjoint variable condition, and of equsal 2422 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 271	. . 3 ⊢ ((𝑥 = 𝑦 → 𝜑) ↔ (𝑥 = 𝑦 → 𝜓))
3	2	albii 1821	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ ∀𝑥(𝑥 = 𝑦 → 𝜓))
4		equsv 2005	. 2 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜓) ↔ 𝜓)
5	3, 4	bitri 275	1 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) ↔ 𝜓)

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1540

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969

This theorem depends on definitions:  df-bi 207  df-ex 1782

This theorem is referenced by:  equsexvw  2007  equvelv  2033  sb6  2091  sbievwOLD  2100  ax13lem2  2381  reu8  3693  el  5396  asymref2  6084  intirr  6085  fun11  6576  fv3  6862  elirrv  9516  fpwwe2lem11  10566  axprALT2  35293  axreg  35311  axregscl  35312  bj-dvelimdv  37126  bj-dvelimdv1  37127  undmrnresiss  43989  pm13.192  44795