Satallax is a higher-orderautomated theorem prover. Satallax won the THF divisionof CASC-23 in 2011.
Automated Reasoning in Higher Order Logic
Let A be an object. Let φ(a) be a proposition depending on a member a of A. Assume ¬(∀a∈A⋅φ(a)). Then we have ∃a∈A⋅¬φ(a).
The following background is necessary to understand this theorem.
(.) Let A and x be objects. Then x∈A is a proposition.
(.) Let φ be a proposition. Then ¬φ is a proposition.
(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Then "∀a∈A⋅φ(a)" is a proposition.
(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Then "∃a∈A⋅φ(a)" is a proposition.
The following background is necessary for the proof.
(.) Let φ and ψ be propositions. Assume φ and ¬φ. Then we know ψ.
(.) "false" is a proposition.
(.) Let φ be a proposition. Assume ¬φ implies false. Then we know φ.
(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Assume for all a ∈ A φ(a). Then we know ∀a∈A⋅φ(a).
(.) Let A be an object. Let φ(a) be a proposition depending on a member a of A. Let a be a member of A. Assume φ(a). Then we know ∃a∈A⋅φ(a).
It is enough to show ¬(∃a∈A⋅¬φ(a)) implies false. Assume ¬(∃a∈A⋅¬φ(a)). Using ¬(∀a∈A⋅φ(a)), it is enough to show ∀a∈A⋅φ(a). Let a be a member of φ. We will show ¬φ(a) implies false. Assume ¬φ(a). Thus ∃a∈A⋅¬φ(a). Using this and ¬(∃a∈A⋅¬φ(a)), we conclude false.
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