Here are solutions for Assignment 1, as the problems were given to (and further explained to) Ganong's class. Since I don't know particulars for Whiteley's class, Whiteley vict -- er, students should read with caution. I believe that the amount of detail and degree of clarity in these solutions are close to what Walter would want to see you write, on assignments or tests, though.

1 One easily checks that the wff's in (a) and (c) are true on all lines of their respective truth-tables (eight lines for (a), two for (c)), whereas that in (b) is false only when A , B and C are assigned truth values F, F, and T respectively.

REMARK: NO , this is not an acceptable sort of answer for you to give on an assignment or test. I am typing a straight text file on a 286, and do not intend to type out 18 or more lines of truth-tables....

2 Here also I will not type out the trees. What I type below is obviously not intended to be an acceptable answer; it just describes acceptable answers : Each tree consists of the given wff at the "top" of the tree, with diagonal (usually) line segments dangling from the "main" connective of each sentence (one such dangler if the connective is the "unary" connective ¬ (what a misnomer for something that does not CONNECT one thing to another...), two such if the connective is one of the other four binary fellows. The wff(s) at the bottom of the dangler(s) is (are) just the "unnegated" wff in the unary case, and the two fellows connected by the connective one level up, in the binary case. And of course one continues in this fashion till atoms are produced (and then stops...). :-)

3 (a) In my section I explained that two truth-tables are to be regarded as being the same iff they have the same list of truth-value assignments (tva's) on the left, in the standard, or lexicographic, ordering for tva's, and have the same "final" column of truth-values for the wff in question (i.e., what the wff actually is is irrelevant in this problem; only its final column of truth-values matters).

So, there are exactly four tables possible, corresponding to the four possibilities for T's and F's in the final column, in rows 1 and 2 of the two-line table, for a wff that involves exactly one atom. ( T in line 1, T in line 2; T in line 1, F in line 2; etc.)

(b) Simplest is perhaps this:

D has truth values T,F in lines 1,2 respectively.
¬ D ----------- F,T ---------------------.
D v ¬ D -------- T,T ---------------------.
D & ¬ D -------- F,F ---------------------.

(c) The argument is already in my answer to (a).

(d) Yes. E.g., the fourth wff \delta above and \alpha := A & ¬ A are both contradictions. (If you take as defn. of "logically equivalent" that the final columns must be the same, then you must regard both the above wff's as involving both A and D (A being invisible in the first, D being invisible in the second). If, as Ganong said in his class that he prefers to do, you take as the definition the alternative defn. using a biconditional on p. 70 in the book, then you automatically get T's on all four lines of the biconditional \alpha \bicond \delta . (Sorry; cannot type a triple bar on this keyboard.)

Two tautologies would of course work as well.

4 (a) The wff's in (a) and (c) are tautologies; that in (b) is a contingent.

(b) Let's use the alternative defn. on p. 70 of the text. Then we are in each case looking at a biconditional that has a four-line truth-table.

(i) If \alpha, \beta are both tautologies, then they are both T on all four lines of their "joint truth-table". So by defn. of \bicond, the biconditional is true on all four lines. So \alpha, \beta are logically equivalent.

REMARK: If you were asked for such an "informal" argument on a test, you would be expected to write about this much detail, and expected to write correct, clear English, as here.

(iii) Similar to (i) -- both wff's are false on all four lines of the joint truth-table, hence the biconditional is again true on all four lines.

(ii) Suppose \alpha, \beta are contingent. Pick a tva \tau for B, for which \beta is false, and a tva \sigma for A, for which \alpha is true. Then the tva (\sigma, \tau) in the four-line table for the biconditional makes the biconditional false. So \alpha, \beta are not logically equivalent.