SEMANTICS
Answers to assignment 15

Every cat hates some mouse

Model <A, F>
A = {a, b, c, d}
F(cat) = {a, b}
F(mouse) = {c, d}
F(hates) = {<a, c>, <b, d>}

Derivation Tree 1
Every cat hates some mouse 

				cat			v1 hates some mouse

						mouse		v1 hates v2

							v1		hates v2

								hates		v2

[[mouse]]M, g(c) = 1 and [[v1 hates v2]]M, gv1av2c = 1. Therefore, by the immediately preceding semantic rule [[v1 hates some mouse]]M, gv1a = 1.
Similarly, [[mouse]]M, g (d) = 1 and [[v1 hates v2]]M, gv1bv2d = 1. Therefore, by the immediately preceding semantic rule [[v1 hates some mouse]]M, gv1b = 1.
Now a and b are the only members of A that are also members of [[cat]]M, g. Therefore, by the penultimate semantic rule cited above, [[every cat hates some mouse]]M, g = 1.
Since g is an arbitrary value assigment, it follows that [[every cat hates some mouse]]M, g = 1.

Derivation Tree 2

Every cat hates some mouse 

				mouse			every cat hates v2

						cat		v1 hates v2

							v1		hates v2

								hates		v2

[[mouse]]M, g(c)=1, but [[every cat hates v2]]M, gv2c=0 because [[v1hates v2]]M, gv2cv1b = 0. Similarly, though [[mouse]]M, g(d) = 1, [[every cat hates v2]]M, gv2d = 0, because [[v1 hates v2]]M, gv2dv1a = 0. Consequently, [[every cat hates some mouse]]M, g = 0. Since, we have shown this to be the case for an arbitrary g, it follows that [[every cat hates some mouse]]M = 0, i.e. the two derivation trees have distinct truth conditions.