Theorem speivw 2000

Description: Version of spei 2432 with a disjoint variable condition, which does not require ax-13 2410 (neither ax-7 2035 nor ax-12 2219). (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 1999	1 ⊢ ∃𝑥𝜑

Syntax hints:   → wi 4   ↔ wb 209  ∃wex 1806

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-6 1994

This theorem depends on definitions:  df-bi 210  df-ex 1807

This theorem is referenced by:  elirrvOLDOLD  9561  bnj1014  35294  eusnsn  47687