Metamath Proof Explorer

Theorem frege48

Description: Closed form of syllogism with internal disjunction. If ph is a sufficient condition for the occurrence of ch or ps and if ch , as well as ps , is a sufficient condition for th , then ph is a sufficient condition for th . See application in frege101 . Proposition 48 of Frege1879 p. 49. (Contributed by RP, 24-Dec-2019) (Proof modification is discouraged.)

		Ref	Expression
	Assertion	frege48	$⊢ (φ \to (\neg ψ \to χ)) \to ((χ \to θ) \to ((ψ \to θ) \to (φ \to θ)))$

Proof

Step	Hyp	Ref	Expression
1		frege47	$⊢ (\neg ψ \to χ) \to ((χ \to θ) \to ((ψ \to θ) \to θ))$
2		frege23	$⊢ ((\neg ψ \to χ) \to ((χ \to θ) \to ((ψ \to θ) \to θ))) \to ((φ \to (\neg ψ \to χ)) \to ((χ \to θ) \to ((ψ \to θ) \to (φ \to θ))))$
3	1 2	ax-mp	$⊢ (φ \to (\neg ψ \to χ)) \to ((χ \to θ) \to ((ψ \to θ) \to (φ \to θ)))$