Metamath Proof Explorer


Theorem exlimih

Description: Inference associated with 19.23 . See exlimiv for a version with a disjoint variable condition requiring fewer axioms. (Contributed by NM, 10-Jan-1993) (Proof shortened by Andrew Salmon, 13-May-2011) (Proof shortened by Wolf Lammen, 1-Jan-2018)

Ref Expression
Hypotheses exlimih.1 ⊢ ( 𝜓 → ∀ 𝑥 𝜓 )
exlimih.2 ⊢ ( 𝜑 → 𝜓 )
Assertion exlimih ( ∃ 𝑥 𝜑 → 𝜓 )

Proof

Step Hyp Ref Expression
1 exlimih.1 ⊢ ( 𝜓 → ∀ 𝑥 𝜓 )
2 exlimih.2 ⊢ ( 𝜑 → 𝜓 )
3 1 nf5i ⊢ Ⅎ 𝑥 𝜓
4 3 2 exlimi ⊢ ( ∃ 𝑥 𝜑 → 𝜓 )