"                                         Homework 2

1.3:

6. (a) NOT ((q IMPLIES NOT p) AND (p IMPLIES r OR q))

(b) NOT (NOT (r OR p IMPLIES q) IMPLIES q AND p)

Both are wffs.

8. (a) It's a wff:                    IMPLIES
                                         /\
                                        /  \
                                       /    \
                                      /   IMPLIES
                                     /       /\
                                    AND     r  s
                                     /\
                                    p  \
                                        \
                                        NOT
                                         |
                                         OR
                                         /\
                                        p  \
                                            \
                                            NOT
                                             |
                                             q

(b) Not a wff: p OR q AND s is ambiguous. But it can be fixed by adding parentheses.
There are two possible ways.

(c) Not a wff: adjacent OR AND can't be parsed.

1.4:

3. (c) & (d) 
p | q | r | r OR p | r OR p IMPLIES q | NOT (r OR p IMPLIES q) | q AND p |
--------------------------------------------------------------------------
T | F | T |   T    |        F         |           T            |    F    |
F | T | F |   F    |        T         |           F            |    F    |


NOT (r OR p IMPLIES q) IMPLIES q AND p | NOT (NOT (r OR p IMPLIES q) IMPLIES q AND p)
-------------------------------------------------------------------------------------
                  F                    |                         T
                  T                    |                         F

6. (a) & (b) p | q | p*q | (p*p)*(q*q)
            --------------------------
             F | F |  F  |      F
             F | T |  T  |      F
             T | F |  T  |      F
             T | T |  F  |      F

(c) Same as BOTTOM (always F).

(d) Yes: exclusive or (XOR).

12. (b) q = F, r = T, p doesn't matter.

(e) p = T, q = F, r = F
.

Additional problem 1
Case NOT i:
Psi = NOT Chi for some wff Chi, and proof looks like

1. ...
.
.
.
     ____________________________________________________
i.   |Chi                                               | assumption
     |.                                                 |
     |.                                                 |
     |.                                                 |
j.   |BOTTOM                                            |
     |__________________________________________________|
.
.
.
n.  NOT Chi                                               NOT i i-j

Consider the proof from lines 1-j without the box and the justification of line
i changed to premise. This is a proof of BOTTOM from premises Phi1,...,Phik,Chi.
Since j < n, by the induction assumption, Phi1,...,Phik,Chi semantically entail BOTTOM.
That is, when Phi1,...,Phik,Chi are all T, so is BOTTOM. But BOTTOM is always F, and
therefore Phi1,...,Phik,Chi can never all be T.

Thus when Phi1,...,Phik are all T, Chi must be F. i.e., NOT Chi is T. This means
Phi1,...,Phik semantically entail NOT Chi = Psi, so we're done with this case.

Case BOTTOM e:
Proof looks like

1. ...
.
.
.
i. BOTTOM
.
.
.
n. Psi                                                     BOTTOM e i

Consider the proof from lines 1-i. This is a proof of BOTTOM from premises Phi1,...,Phik.
Since i < n, by the induction assumption, Phi1,...,Phik semantically entail BOTTOM.
That is, when Phi1,...,Phik are all T, so is BOTTOM. But BOTTOM is always F, and
therefore Phi1,...,Phik can never all be T.

Thus when Phi1,...,Phik are all T (which can never happen), Psi is T. This means
Phi1,...,Phik semantically entail Psi, so we're done with this case.

Additional problem 2
Case root of parse tree is labeled AND (this is very similar to the case in class,
where the root is labeled OR):

Psi = Phi AND Chi for some wffs Phi, Chi. So the parse tree of Psi has two subtrees
that are the parse trees of Phi and Chi, and both of these subtrees have height < n.

Subcase: the truth assignment makes Phi, Chi both F.
Then Phi^ = NOT Phi, Chi^ = NOT Chi, and
Psi^ = NOT Psi = NOT (Phi AND Chi).

By the induction assumption,
there is a proof of Phi^ from premises Phi1^,...,Phik^, and
there is a proof of Chi^ from premises Phi1^,...,Phik^.

There is a proof of NOT (Phi AND Chi) from premises NOT Phi,NOT Chi. You should be able
to do it. So altogether there is a proof of Psi^ = NOT (Phi AND Chi) from premises
Phi1^,...,Phik^. This finishes the first subcase.

There are three more subcases:
truth assignment makes Phi F, Chi T
truth assignment makes Phi T, Chi F
truth assignment makes Phi T, Chi T

They have the same form as the first subcase above; in fact the second and third are so
similar that you could just do one of them and say that the other is essentially the same.