Theorem speivw 2006

Description: Version of spei 2425 with a disjoint variable condition, which does not require ax-13 2403 (neither ax-7 2041 nor ax-12 2215). (Contributed by BJ, 31-May-2019.)

Hypotheses

Ref	Expression
speivw.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
speivw.2	⊢ 𝜓

Assertion

Ref	Expression
speivw	⊢ ∃𝑥𝜑

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)   𝜓(𝑥, 𝑦)

Proof of Theorem speivw

Step	Hyp	Ref	Expression
1		speivw.1	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
2	1	biimprd 251	. 2 ⊢ (𝑥 = 𝑦 → (𝜓 → 𝜑))
3		speivw.2	. 2 ⊢ 𝜓
4	2, 3	speiv 2005	1 ⊢ ∃𝑥𝜑

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∃wex 1812

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-6 2000

This proof depends on definitions:  df-bi 210  df-ex 1813

This theorem is used by:  elirrvOLDOLD  9574  bnj1014  35457  eusnsn  47901