Logic Quiz Study Guide for Quiz 1
I will walk through some of the exercises in detail to illustrate and remind you of important concepts.
Symbolization
If Z remembered to do his chores then things are clean but not neat. If he forgets, then things are neat but not clean. Therefore, things are either neat or clean, but not both.
I find it helpful to put this in the form of a premise-conclusion argument while translating:
(1) If Z then C and not-N
(2) If not-Z, then N and not-C
(3) Therefore, N or C and not N and C
Note I’ve left some scope ambiguities, e.g. “not N and C” in line (3) could be read as either “not-(N and C) or “not-N and C.” The former is the correct reading, but I wanted to walk you through these ambiguities explicitly. Consider next, where I use “~” for negation:
(1) Z -> (C /\ ~N)
(2) ~Z -> (N /\ ~C)
(3) (N \/ C) /\ ~(N /\C)
You can see whether (3) follows by examining the premises. We know either Z is true or ~Z is true. If Z is true then C is, from which it follows that (N \/ C) is. If ~Z is true, then N is, from which it follows that (N \/ C) is. Hence, in either case (N \/ C) is true. Hence, we have the left side. Similarly, if Z is true, then ~N is, from which it follows that ~N \/ ~C is, from which it follows ~(N /\C) is. If ~Z is true, then ~C is, from which it follows that ~N \/ ~C is, from which it follows ~(N /\ C) is. Hence, in either case ~(N /\ C) is true. Hence, we have the right side of (3).
Note too that line (2) could be translated as (N /\ ~C) unless Z, i.e. “Things are neat but not clean unless Z remembered to do his chores.” This is because:
P unless Q =
P, if not Q =
If not Q, then P
Speaking of conditionals, remember that “if” not preceded by “only” introduces the antecedent of a material conditional. The antecedent of the preceding unless statement is complex, since it has a negation inside it. You can read that as: P, if [not-Q], where the antecedent is everything in the brackets.
This is distinct from “only if” which introduces the consequent of an expression. That is:
P only if Q =
If P then Q
Putting these observations about “if” and “only if” together, we can define “if and only if” as, where I bracket each side of the conjunction for clarity:
P if and only if Q =
[P, if Q] and [P only if Q]
Which is to say – and this is important – that conjunction comes first in priority here. That is, you bracket the conditionals after you separate the conjunction sides. The above is equivalent to:
[Q -> P] /\ [P -> Q]
Semantic Concepts
Suppose A and B are logically equivalent. That means they always have the same truth-table values. For example:
P->Q: T F T T
~Q -> ~P: T F T T
Suppose then you have:
(P -> Q) <-> (~Q -> ~P)
This is equivalent to:
[(P -> Q) -> (~Q -> ~P)] /\ [(~Q -> ~P) -> (P -> Q)]
Look at just the left side:
[(P -> Q) -> (~Q -> ~P)]
This claims that whenever (P->Q) is true then so is (~Q->~P). This must be the case since they always have the same truth value assignment. Hence, the left side must be true. Similarly for the right. But then both sides of the biconditional are always true when the sides are logically equivalent. Hence, we’ve a tautology. A simpler example is:
P <-> P
Clearly this is always true. When P is true so is P, hence P<->P is true. When P is false, so is P, hence P<->P is (biconditionals are true just in case both sides have the same truth value). Hence, P<->P is a tautology.
Suppose next that we have:
(1) R /\ S
(2) T
(3) P <->P
We know the conclusion is a tautology. We are examining whether (3) follow from (1) and (2). We check validity by examining whether there are any cases in which the premises are all true and the conclusion false. But note here, the conclusion is a tautology, and so is never false.
Now let’s return to our logically equivalent A and B. Let’s consider two arguments:
(1) A
(2) Therefore, B
And:
(1) B
(2) Therefore, A
And examine the validity of each. Note, if A and B are logically equivalent, then they have the same truth assignments. So, whenever A is true/false, so is B. Recall to the definition of validity:
An argument R is valid if and only if when all R’s premises are true, so is R’s conclusion
As stated, this is a biconditional. That means if R is a valid argument, then when all R’s premises are true then so is R’s conclusion. Similarly, when all R’s premises are true then so is R’s conclusion, R is valid.
We have some options with respect to our pair of arguments above. Suppose A is true. Then so is B. Then it’s true that if A is true then so is B. Hence, we’ve found no counterexample to the validity of the first argument. We have more work to do of course. A is either true or false. Suppose A is false. Well then it doesn’t matter what B is (B is false of course) for the sake of validity. You can only have a counterexample to validity (and so an invalid argument) when all premises are true. Since A is false, we count this first argument as valid. I’ll let you make the similar observations about the second argument.
Partial Truth Tables
(1) A -> B: T F T T
(2) B: T F T F
(3) A: T T F F
We’re checking if (3) follows by constructing truth tables. I find it somewhat clearer to write the truth tables in the preceding way, since it illustrates it’s an argument we’re considering. To check whether (3) follow we look for counterexamples to validity, i.e. cases where all the premises are true and the conclusion is false. Observe there are two rows where both premises are true. In the second, we see that A is false. Hence, this is not a valid argument.
Now, let’s check whether ~(A /\ ~B), B -> ~A, and ~B are jointly satisfiable. Consider:
~(A /\ ~B): T F T T
B -> ~A: F T T T
~B: F T F T
Since there is one row in which all assignments are true, this set is jointly satisfiable.
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[What follows is priming for later in the quarter, so you don’t to worry about it now if you’d prefer to focus for the quiz]
You can make the preceding easier by observing the first expression is equivalent to: A -> B, so you have:
A -> B
B -> ~A
~B
From the first two you can argue:
(1) A -> B
(2) B -> ~A
(3) Therefore, A -> ~A (contradiction!)
But, if we assume both A and B are false, then what we have expressions of the form:
~A -> ~B
~B -> A
B
From which we might argue:
(1) ~A -> ~B
(2) B
(3) Therefore, ~~B
(4) Therefore, ~~A
(5) Therefore, A
But nothing much else follows of interest, and certainly no contradiction.
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