I think the anwser should be B). At first, I thought of A) as the correct answer, but on a second thought, in the statement "The Racket identifier may not be used if it has not been defined." did not use only if. So, this statement is equivalent to "if it has not been defined, then the Racket identifier may not be used.
Yes. $\neg d\to\neg u$ says "Not $u$ if not $d$", and also "If not $d$, then not $u$."
I believe the answer should be E), since the statement "If Mary-Kate and Ashley are identical twins, then either they both have blue eyes or neither has blue eyes." translates to the conditional:
$t→((m∧a)∨¬(m∨a))$, and the consequent does not correspond to any conditional/biconditional statements in A) through D). So E) may be the answer.
No. The consequent is saying that the truth values for $m$ and $a$ are equal, which is a bicondition. $${\quad(m\wedge a)\vee\neg(m\vee a) \\\equiv (m\wedge a)\vee(\neg m\wedge \neg a)\\\equiv (m\vee\neg a)\wedge(\neg m\vee a)\\\equiv (a\to m)\wedge(m\to a)\\\equiv m\leftrightarrow a}$$
According to my textbook, the negation law states that
$p∨¬p≡T$ and $p∧¬p≡F$. So I can use first use associative law to get
$p∧((a∨b)∧¬(a∨b))$ then apply negation law to get $p∧F$. So the answer is D)
Indeed, that is so.
Also, what does the F and T mean in the negation law. I mean $p∧¬p$, does this is a tautological falsehood (contradiction) right?
T : true, F : false. $\phi\vee\neg\phi$ is true, $\phi\wedge\neg\phi$ is false.
Indeed since they are true and false reguardless of the value of $\phi$, they are a tautology and contradiction respectively.
Best Answer
Associativity applies only when the connectives involved are exclusively $\land$ or exclusively $\lor$:
$$p \land q \land r \equiv (p \land q)\land r \equiv p \land (q\land r)$$
$$p \lor q \lor r \equiv (p \lor q)\lor r \equiv p \lor (q\lor r)$$
Because of associativity of $\lor$ and $\land$, parentheses are not necessary to define expressions like those above.
Your statement, however:
$$p \land \lnot q \lor q \land \lnot r \lor \lnot p \lor r \tag{given}$$
has mixed connectives, and so associativity does not apply across all possible groupings.
Please note: as stated, your (given) expression is not well-defined without parentheses. That is, without parentheses, it is ambiguous; it can be read any number of ways, most of which are not equivalent. Does it mean connect from left to right?:
$$(((((p\land \lnot q) \lor q) \land) \lnot r)\lor\lnot p) \lor r\;?\tag{1}$$
Or does it mean this?
$(p \land \lnot q) \lor (q \land \lnot r) \lor (\lnot p \lor r)\;?\tag{2}$
or any number of other possible ways of grouping with parentheses?
In general, when you have an expression like $(2)$ above, you need to apply the Distributive Laws to distribute over another connective:
For example $$p \land (q \lor r) \equiv (p \land q) \lor (p \land r)$$ $$p \lor (q\land r) \equiv (p \lor q) \land (p\lor r)$$