Our predicates are and S, where (x, y, z) is to be interpreted as \'the product

Question

Our predicates are and S, where (x, y, z) is to be interpreted as 'the product of x and y is z' (satisfied by all ordered triples of members of the underlying domain ("Universe of Discourse") U whose third term (i.e., the term in the third position) is equal to the product of the first two terms (i.e., the terms in the first and second positions)), and S(w, x, y, z) is to be interpreted 'the sum of w and x is equal to the sum of y and z' (i.e., w + x = y + z, satisfied by all ordered quadruples of elements of U the sum of whose first two terms is numerically identical to the sum of whose last two terms). Either by enumeration or verbal or symbolic description, identify the truth values or sets (cither of elements, ordered pairs of elements, ordered triples of elements, ordered quadruples of elements, etc. of the specified domain of U) defined by the following wffs, and give a brief indication of your reasoning. (Forall y) (x, y, z) - U = N = {0, 1, 2, 3 ...} (x)S(8, x, y, z) - U = N = {0, 1, 2, 3 ....} (Forall w)(Forall y)S(w, x, y, z) - U = N {0, 1, 2, 3, ....}

Explanation / Answer

1) The expression means that 'for all y, x * y = z' where x,y,z are elements of N =[0,1,2,...}.

Let first y = 0 (fixed) and we give different values for x.

Then we have 0*0=0, 1*0= 0, 2*0=0,3*0=0.....and so on.Thus the triplets are (0,0,0), (1,0,0),(2,0,0,),(3,0,0)....

Similarly, when y= 1(fixed) and we give different values for x.

Then we have 0*1=0,1*1=1,2*1=2,3*1=3......and so on. Thus the triplets are (0,1,0),(1,1,1),(2,1,2),(3,1,30......

Next giving y = 2 with varying x we obtain the triplets (0,2,0),(1,2,2),(2,2,4),(3,2,6)........and so on.

2) This expression means that 'there exists x such that 8+x = y+z'

Consider x= 0 in N={0,1,2,3,...}.

Then, 8+0 = 0+8 or 1+7 or 2+6 or 3+5 or 4+4 or 5+3 or 6+2 or 7+1 .

Thus, 8+x = y+ z corresponds to the quadruples (8,0,0,8),(8,0,1,7),(8,0,2,6),(8,0,3,5),(8,0,4,4,),(8,0,5,3),(8,0,6,2,)(8,0,7,1) for x= 0 alone !

This can be done with any value for x in N= {0,1,2,...}

3)This expression means that 'for all w and for all x, there exists y such that w+x = y+z'.

Let w=x= 0 in N=[0,1,2,..}

Then y= z = 0 so that 0+0 = 0+ 0. Thus the quadruple (0,0,0,0)

Next let w=x= 1

Then y = z =1 so that 1+1=1+1, also y=2, z=0 so that 1+1=2+0, again y=0,z=2 so that 1+1=0+2.

Thus the quadruples (1,1,1,1),(1,1,2,0),(1,1,0,2).

This can be repeated with w=x= 2 to get the quadruples (2,2,2,2)(2,2,4,0)(2,2,0,4)(2,2,1,3)(2,2,3,1) and so on.

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