MTH 714 Lecture Notes - Lecture 10: Logical Consequence, Unary Function, Ryerson University
9 Dec 2015
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Problem set #4: chapters 9, 10 and 11. X y( zp(z) u( q(x, u) vq(y, v))) X y z u v(p(z) ( q(x, u) q(y, v))) X y(p(f (x, y) ( q(x, g(x, y)) q(y, h(x, y)))) P(b) q(c) r(a: exercise 9. 5. , xn) is satis able in a model with only one element it is satis able in general. ( ) now, suppose x1 . , am}) where ri are relations on d interpreting predicate symbols and aj are elements which interpret constant symbols in the formula. We want to show that the formula is true in some interpretation whose domain has only one element. Fix one element of d and denote it b. , xn) expresses the fact that some property holds for all elements x1, . , xn of d, in particular, it will be true when all xi = b and we interpret all constant symbols as that same element b.
