Write expressions w/out quantifiers (convert to AND/OR expressions)

Question

2. A universe contains the three individuals $a,b$, and $c$. For these individuals, a predicate $Q(x,y)$ is defined, and its truth values are given by the following table: \begin{array}{c|ccc} x\backslash y&a&b&c\\\hline a&T&F&T\\ b&F&T&F\\ c&F&T&T \end{array} Write each of the following expressions without quantifiers (i.e. covert them to expressions with ANDs and ORs or both) and then evaluate the expression.
   a) $\forall x\,\exists y\,Q(x,y)$
   b) $\forall y\,Q(y,b)$
   c) $\forall y\,Q(y,y)$

I found this post which says to convert all $\exists$ to OR and all $\forall$ to AND. However, my problem has 2 variables (instead of one, like in the aforementioned example). I'm not sure how to handle that.

Source.

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My attempts:

a) $[P(a,a)\lor P(b,a)\lor P(c,a)]\land [P(b,a)\lor P(b,b)\lor P(b,c)]\land [P(c,a)\lor P(c,b)\lor P(c,c)]$

b) $P(a,b)\land P(b,b)\land P(c,b)$

c) $P(a,b)\land P(a,c)\land P(b,a)\land P(b,c)\land P(c,a)\land P(c,b)$

I am unsure how to evaluate the expressions..

To evaluate the expressions, I just need to match the "coordinates" with their respective T/F then evaluate whether or not the whole expression is T/F from there. Correct?

Answer 1

However, my problem has 2 variables (instead of one, like in the aforementioned example). I'm not sure how to handle that.

a) $[P(a,a)\lor P(b,a)\lor P(c,a)]\land [P(b,a)\lor P(b,b)\lor P(b,c)]\land [P(c,a)\lor P(c,b)\lor P(c,c)]$

Almost as you did; but not quite (you muddled up the first grouping). (Also, mind your Pees and Queues.)

When in doubt, write it out, just one step at a time. $$\begin{align} &\qquad \forall x~\exists y~Q(x,y) \\ & \equiv \big(\exists y~Q(a, y)\big)\wedge\big(\exists y~Q(b, y)\big)\wedge \big(\exists y~Q(c, y)\big) \\ & \equiv \big(\exists y~Q(a, y)\big)\wedge\big(\exists y~Q(b, y)\big)\wedge\big(Q(c,a)\vee Q(c,b)\vee Q(c,c)\big) \\[1ex] & \equiv \big(Q(a,a)\vee Q(a,b)\vee Q(a,c)\big)\wedge\big(Q(b,a)\vee Q(b,b)\vee Q(b,c)\big)\wedge\big(Q(c,a)\vee Q(c,b)\vee Q(c,c)\big) \\[1ex] & \equiv \big(\mathrm T\vee \mathrm F\vee \mathrm T\big)\wedge\big(\mathrm F\vee \mathrm T\vee \mathrm F\big)\wedge\big(\mathrm F\vee \mathrm T\vee \mathrm T\big) \\ & \equiv (\mathrm T)\wedge(\mathrm T)\wedge(\mathrm T) \\ & \equiv \mathrm T \end{align}$$

b) $P(a,b)\land P(b,b)\land P(c,b)$

$\forall y\;Q(y,b) = Q(a,b)\land Q(b,b)\land Q(c,b) \quad\checkmark$

c) $P(a,b)\land P(a,c)\land P(b,a)\land P(b,c)\land P(c,a)\land P(c,b)$

$\forall y\;Q(y,y) = Q(a,a)\land Q(b,b)\land Q(c,c) \quad$ all $y$ within each term are replaced with the same instance.

Answer 2

$$\forall y ~ Q(y, y)$$ $$P(a,b)\land P(a,c)\land P(b,a)\land P(b,c)\land P(c,a)\land P(c,b)$$

$$\forall y ~ Q(y, y)$$

Should be

$$Q(a, a) \land Q(b, b) \land Q(c,c)$$

Otherwise you are correct. Edit: As tunococ pointed out, in (a), $[P(a,a)\lor P(b,a)\lor P(c,a)]$ has a mistake not present in the other two terms.

To evaluate the expressions, I just need to match the "coordinates" with their respective T/F then evaluate whether or not the whole expression is T/F from there. Correct?

Yes.