SOLUTIONS TO ASSIGNMENT 3Question 1(20)(i) To prove that the formula∃x∀yp(x, y)→ ∀y∃xp(x, y) is valid, we must show that it is true inall interpretations. We do this by contradiction.Say there is an interpretationIin which the formula is false, i.e.vI(∃x∀yp(x, y)→ ∀y∃xp(x, y)) =F. ThenvI(∃x∀y(p(x, y)) =TandvI(∀y∃xp(x, y)) =F(by the truth values of implication).ThenvσI(∀yp(x, y)) =Tfor some assignmentσIandvσ0I(∃xp(x, y)) =Ffor some assignmentσ0I(both by Theorem 7.22).According to Definition 7.18,σImaps the only free variable in∀yp(x, y), namelyx, to a valued∈D.SincevσI(∀yp(x, y)) =T, according to Definition 7.19vσI[y←ei](p(x, y)) =Tfor allei∈D.Also according to Definition 7.18,σ0Imaps the only free variable in∃xp(x, y), namelyy, to avaluee∈D. Sincevσ0I(∃xp(x, y)) =F, according to Definition 7.19vσ0I[x←dj](p(x, y)) =Fforalldj∈D.But this is a contradiction because by Definition 7.19 the ordered pair (d, e) must be inRp(the relation assigned topinI) to comply withvσI[y←ei](p(x, y)) =Tfor somed∈Dand allei∈D, but (d, e) may not be inRpto comply withvσ0I[x←dj](p(x, y)) =Ffor somee∈Dandalldj∈D.So our assumption, namely that there is an interpretationIin which the formula is false, isincorrect, and we conclude that the formula is valid.BTW, as proved in Question 2, the converse of this formula is not valid. See if you can workout why a validity proof similar to the above won’t work.(ii) To prove that∃x(p(x)→q(x))↔(∀xp(x)∧ ∀x¬q(x)) is unsatisfiable, we must show that it isfalse in all interpretations.LetIbe an arbitrary interpretation of the formula.vI(∃x(p(x)→q(x))) =FiffvσI(p(x)→q(x)) =Ffor all assignmentsσI(by the semantics of∃)iffvσI(p(x)) =TandvσI(q(x)) =Ffor all assignmentsσI(by the truth values for implication)iffvσI(p(x)) =TandvσI(¬q(x)) =Tfor all assignmentsσI(by the truth values for negation)iffvI(∀xp(x)) =TandvI(∀x¬q(x)) =T(by the semantics of∀)iffvI(∀xp(x)∧ ∀x¬q(x)) =T(by the truth values of conjunction)SincevI(∃x(p(x)→q(x)) =FiffvI(∀xp(x)∧ ∀x¬q(x)) =T, we conclude thatvI(∃x(p(x)→q(x))6=vI(∀xp(x)∧ ∀x¬q(x)), and thus that