Questions on 3.5?
Here’s Example 3.5.7 (a direct proof), done in prose:
Claim: p -> r, q -> s, p-or-q => r-or-s.
Direct proof: Suppose p->r, q->s, and p-or-q are all true.
Since p-or-q is true, p is true or q is true.
Case 1: If p is true, then the truth of p combined with the
truth of p->r implies the truth of r, which implies the truth
of r-or-s.
Case 2: If q is true, then the truth of q combined with the
truth of q->s implies the truth of s, which implies the truth
of r-or-s.
Since in both cases we have deduced the truth of r-or-s, the
truth of the claim follows.
Here’s Example 3.5.17 (an indirect proof), done in prose.
Claim: a->b, not-(b or c) => not-a.
Indirect proof: Suppose a->b and not-(b-or-c). Suppose
furthermore (for purposes of contradiction) that not-a is
false; that is, suppose a is true.
Since a is true and a->b is true, b is true.
This implies that b-or-c is true, implying in turn that not-(bor-c) is false.
But since we supposed that not-(b-or-c) is true, this is a
contradiction.
The truth of the claim follows.
Group work: 3.5.2
We saw in class that (q-and-(not-q)) => p, regardless of the
nature of the propositions p and q. That’s because the
antecedent is always false, and we’ve defined an
implication to be (vacuously) true when the antecedent is
false. Or, putting it differently, the truth set of q-and-(not-q)
is the empty set, which must be a subset of the truth set of
p.
(Here’s a proof in the style of Doerr and Levasseur: Since
(p or q) can be deduced from the hypothesis q and since
not-q is one of the given hypotheses, the disjunctive
simplification rule, applied in the form “(p or q) and (not q)
=> not p”, gives us the conclusion p.)
In a similar way, we can check that q => (p-or-(not-p)),
regardless of the nature of the propositions p and q. That’s
because the consequent is always true, and an implication is
true when the consequent is true. Or, putting it differently,
the truth set of (p-or-(not-p)) is the (whole) universe, and
the truth set of q must be a subset of the universe.
Group work: 3.5.6:
Let p = “x does well in discrete math”, q = “x studies hard”,
r = “x skips classes”, and s = “x does well in courses”.
Then our deduction has the form
p -> q, s -> not-r, q -> s => (p -> (not-r))
This is a valid deduction.
Proof: Suppose p -> q, s -> not-r, and q -> s are all true.
To prove that p -> (not-r) is true, suppose furthermore that
p is true.
Since p is true and p -> q is true, q is true.
Since q is true and q -> s is true, s is true.
Since s is true and s -> not-r, not-r is true.
Thus we have shown that p implies not-r, as was to be
shown.
Other questions on 3.5?
Questions on 3.6?
Propositions over a universe
p => q is equivalent to the assertion that T_p is a subset of
T_q.
Also, p <==> q is equivalent to the assertion that T_p
equals T_q.
Suppose p is the proposition “x is positive” and q is the
proposition “x has a square root”. Then T_p = (0, infinity)
while T_q = [0, infinity). Since T_p is a subset of T_q but
not vice versa, p => q but not vice versa.
Group work: 3.6.2:
T_p = {2,3,5,7,11,13,17,19,23,29,31}
T_q = {1,3,9,27,81,…}
T_r = {1,3,9,27}
T_r is a subset of T_q, so r => q.
Group work: 3.6.7
Other questions on 3.6?
Collect section summaries and homework.