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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 )