Question 3

For each of the modal arguments below, show that they are invalid by providing a counterexample. Show the invalidity of the modal argument by providing the extensions of R(x,y) and P(x), R(x), and Q(x) in a way that makes all the formulas to the left of true and the formula on the right of false at a. Ignore the a=a: that is always true anyway.

NOTE: All of the objects must be natural numbers, i.e., 0,1,2,3 . . . The extension of a unary predicate is just a list of numbers, e.g., 1,2,4 . . .

The extension of a name is just a number, e.g., 1.

To enter the extensions of relations enter them like: , That would mean the extension of, say, R(x,y), contains those two pairs.

Remember, try to keep your domains small.

(PAQETO((PV-vo-Pv ) FE.3.1 3 pts a=a Domain:0 RODI PUT QU: + + a: 0 Submit O(PAO- POOP OOP FE.3.2 3 pts ara Domain:0 ROJ: POD a: 0 ++ Submit Check O-PAQ-O-PA OOR) EKTB (-PA-R) - 0-PA- FE.3.3 3 pts a=a Domain:0 ROJ: POI Ruf QUE a: 0 Submit (PAQETO((PV-vo-Pv ) FE.3.1 3 pts a=a Domain:0 RODI PUT QU: + + a: 0 Submit O(PAO- POOP OOP FE.3.2 3 pts ara Domain:0 ROJ: POD a: 0 ++ Submit Check O-PAQ-O-PA OOR) EKTB (-PA-R) - 0-PA- FE.3.3 3 pts a=a Domain:0 ROJ: POI Ruf QUE a: 0 Submit