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Solved Examples on Symbolic Logic: Symbols into Predicate Logic

Solve all questions.
Show all work.
You may use any of the Laws of Logical Equivalences and/or Logical Equivalences for Propositional Logic and Predicate Logic.
For any law of logical equivalence that you use, please indicate the law or the logical equivalence number.

(1.)

$ LHS \\[3ex] \neg[\forall x (p(x) \lor q(x))] \\[3ex] \equiv \neg \forall x \; \neg(p(x) \lor q(x)) \\[3ex] *** \\[3ex] \neg \forall x \equiv \exists x ...Number(11.)...De\:\:Morgan's\:\:Law \\[3ex] \neg(p(x) \lor q(x)) \equiv \neg p(x) \land \neg q(x) ...Number(1.)...De\:\:Morgan's\:\:Law \\[3ex] *** \\[3ex] \equiv \exists x(\neg p(x) \land \neg q(x)) \\[3ex] = RHS $

(2.)

$ LHS \\[3ex] \neg[\forall x (p(x) \land q(x))] \\[3ex] \equiv \neg \forall x \; \neg(p(x) \land q(x)) \\[3ex] *** \\[3ex] \neg \forall x \equiv \exists x ...Number(11.)...De\:\:Morgan's\:\:Law \\[3ex] \neg(p(x) \land q(x)) \equiv \neg p(x) \lor \neg q(x) ...Number(2.)...De\:\:Morgan's\:\:Law \\[3ex] *** \\[3ex] \equiv \exists x(\neg p(x) \lor \neg q(x)) \\[3ex] = RHS $

(3.) $\forall x\; \forall y\;(x \lt y)$

$\forall x\; \forall y\;(x \lt y)$

(I.) For all positive integers $x$ and $y$, $x \lt y$
What about $3$ and $2$?
$3 \not \lt 2$

(II.) The statement is false.
The truth value is $F$

(4.) $\forall x\; \forall y\;(x \ge y)$

$\forall x\; \forall y\;(x \ge y)$

(I.) For all positive integers $x$ and $y$, $x \ge y$
What about $2$ and $3$?
$2 \not \ge 3$

(II.) The statement is false.
The truth value is $F$

(5.) $\forall x\; \exists y\;(x \gt y)$

$\forall x\; \exists y\;(x \gt y)$

(I.) For every positive integer $x$, there exists at least one positive integer $y$ such that $x \gt y$
If we have $x = 3$, we have $y = 2$ such that $3 \gt 2$
If we have $x = 10$, we have $y = 7$ such that $10 \gt 7$

But, what if we have $x = 1$?
Keep in mind that $0$ is NOT a positive integer.
$1$ is a positive integer.
Do we have any positive integer less than $1$?
We do not.

(II.) The statement is false.
The truth value is $F$

(6.) $\forall x\; \exists y\;(x \ge y)$

$\forall x\; \exists y\;(x \ge y)$

(I.) For every positive integer $x$, there exists at least one positive integer $y$ such that $x \ge y$
If we have $x = 3$, we have $y = 2$ such that $3 \ge 2$
If we have $x = 10$, we have $y = 7$ such that $10 \ge 7$

But, what if we have $x = 1$?
Keep in mind that $0$ is NOT a positive integer.
$1$ is a positive integer.
Do we have any positive integer less than or equal to $1$?
Yes, we do.
$1 \ge 1$.
If we have $x = 1$, we have $y = 1$ such that $1 \ge 1$
This is similar to saying that: There is a smallest positive integer.

(II.) The statement is true.
The truth value is $T$

(7.)

$ RHS \\[3ex] \neg[\exists x (\neg p(x) \rightarrow q(x))] \\[3ex] \equiv \neg \exists x \; \neg(\neg p(x) \rightarrow q(x)) \\[3ex] *** \\[3ex] \neg \exists x \equiv \forall x ...Number(12.)...De\:\:Morgan's\:\:Law \\[3ex] \neg p(x) \rightarrow q(x) \equiv p(x) \lor q(x) ...Number(4.) \\[3ex] \therefore \neg(\neg p(x) \rightarrow q(x)) \equiv \neg (p(x) \lor q(x)) \\[3ex] \neg (p(x) \lor q(x)) \equiv \neg p(x) \land \neg q(x) ...Number(1.)...De\:\:Morgan's\:\:Law \\[3ex] *** \\[3ex] \equiv \forall x(\neg p(x) \land \neg q(x)) \\[3ex] = LHS $

(8.)

$ LHS \\[3ex] \neg[\forall x (p(x) \rightarrow \neg q(x))] \\[3ex] \equiv \neg \forall x \; \neg(p(x) \rightarrow \neg q(x)) \\[3ex] *** \\[3ex] \neg \forall x \equiv \exists x ...Number(11.)...De\:\:Morgan's\:\:Law \\[3ex] p(x) \rightarrow \neg q(x) \equiv \neg p(x) \lor \neg q(x) ...Number(5.) \\[3ex] \therefore \neg(p(x) \rightarrow \neg q(x)) \equiv \neg(\neg p(x) \lor \neg q(x)) \\[3ex] \neg(\neg p(x) \lor \neg q(x)) \equiv \neg \neg p(x) \land \neg \neg q(x) ...Number(1.)...De\:\:Morgan's\:\:Law \\[3ex] \neg \neg p(x) \equiv p(x) ...Number(15.)...Double\:\:Negation\:\:Law \\[3ex] \therefore \neg \neg p(x) \land \neg \neg q(x) \equiv p(x) \land q(x) \\[3ex] *** \\[3ex] \equiv \exists x(p(x) \land q(x)) \\[3ex] = RHS $

(9.)

$ LHS \\[3ex] \neg[\forall x (p(x) \lor (q(x) \lor r(x)))] \\[3ex] \equiv \neg \forall x \; \neg(p(x) \lor (q(x) \lor r(x))) \\[3ex] *** \\[3ex] \neg \forall x \equiv \exists x ...Number(11.)...De\:\:Morgan's\:\:Law \\[3ex] \neg(p(x) \lor (q(x) \lor r(x))) \equiv \neg p(x) \land \neg(q(x) \lor r(x)) ...Number(1.)...De\:\:Morgan's\:\:Law \\[3ex] \neg(q(x) \lor r(x)) \equiv \neg q(x) \land \neg r(x) ...Number(1.)...De\:\:Morgan's\:\:Law \\[3ex] \therefore \neg(p(x) \lor (q(x) \lor r(x))) \equiv \neg p(x) \land \neg q(x) \land \neg r(x) \\[3ex] *** \\[3ex] \equiv \exists x(\neg p(x) \land \neg q(x) \land \neg r(x)) \\[3ex] \equiv \exists x(\neg p(x) \land (\neg q(x) \land \neg r(x))) \\[3ex] = RHS $

(11.)

$ LHS \\[3ex] \neg[\exists x (\neg p(x) \lor (\neg q(x) \lor \neg r(x)))] \\[3ex] \equiv \neg \exists x \; \neg(\neg p(x) \lor (\neg q(x) \lor \neg r(x))) \\[3ex] *** \\[3ex] \neg \exists x \equiv \forall x ...Number(12.)...De\:\:Morgan's\:\:Law \\[3ex] \neg(\neg p(x) \lor (\neg q(x) \lor \neg r(x))) \equiv \neg \neg p(x) \land \neg(\neg q(x) \lor \neg r(x)) ...Number(1.)...De\:\:Morgan's\:\:Law \\[3ex] \neg(\neg q(x) \lor \neg r(x)) \equiv \neg \neg q(x) \land \neg \neg r(x) ...Number(1.)...De\:\:Morgan's\:\:Law \\[3ex] \neg \neg p(x) \equiv p(x) ...Number(15.)...Double\:\:Negation\:\:Law \\[3ex] \neg \neg q(x) \equiv q(x) \\[3ex] \neg \neg r(x) \equiv r(x) \\[3ex] \therefore \neg(\neg q(x) \lor \neg r(x)) \equiv q(x) \land r(x) \\[3ex] \therefore \neg \neg p(x) \land \neg(\neg q(x) \lor \neg r(x)) \equiv p(x) \land q(x) \land r(x) \\[3ex] *** \\[3ex] \equiv \forall x(p(x) \land q(x) \land r(x)) \\[3ex] \equiv \forall x(p(x) \land (q(x) \land r(x))) \\[3ex] = RHS $