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Suggested problems - solutions
Revisiting the rules of inference
Material for this section references College Geometry: A Discovery Approach, 2/e, David C. Kay, Addison Wesley, 2001. In particular, see section 2.1, pp 51-62.
The problems are all from section 2.1.
Problems: 3, 4, 5, 8, 9, 13, 14
#3 Write as an if-then: “Two angles whose measures are each 30◦ are congruent.”
Solution: “If there are two angles whose measures are each 30◦ , then the angles are congruent.” The hypothesis (the “if”) is that there are (given) two angles with measures 30◦ , and the
conclusion (the “then”) is that they are congruent.
#4 Write as an if-then: “Every equilateral triangle is equiangular.”
Solution: “If a triangle is equilateral, then it is equiangular.” The hypothesis is that there is
(given) an equilateral triangle, and the conclusion is that the triangle is equiangular.
#5 Write as an if-then: “The medians of a triangle are concurrent lines.”
Solution: “If some things are the medians of a triangle, then those things are concurrent lines.”
The hypothesis is that you are given (presumably lines which are) the medians of a triangle, and
the conclusion is that they are concurrent.
Note we haven’t defined median of a triangle, or concurrent line for that matter, and you can
still do this. The point is that you don’t even need to know what the terms mean to be able
to restructure a statement as an if-then: “an A is a B” can immediately be written as “if it’s
an A, then it’s a B.” You could just as easily have the statement “The snarfs of a woggle are
farmished” ... and instantly produce “if something is the snarf of a woggle, then it is farmished,”
and go on to identify the hypothesis and conclusion. The logician and author Lewis Carroll
(Alice’s Adventures in Wonderland ) made extensive use of this sort of “illogical logic” to show
the emphasis on structure over meaning.
#8 Consider
Definition: a is said to be b iff p → q
Theorem 1: p → r
Theorem 2: r → q
Prove the following:
Theorem: If p, then a is b.
Proof
Given: p
Prove: a is b
Before filling in the proof, let me mention the “iff” - “if and only if”. This is the biconditional,
and it means we have a logical equivalence - for the purposes of this problem, “a is said to be
b” and “p → q” are exactly the same, and interchangeable. Proving either of those statements
automatically proves the other, and this is reflected in Kay’s setup of the steps. What is proven
is “p → q”; what is concluded is “a is b”. Logically, that’s fine, but it would be a bit better (and
more explicit as to what’s going on) if you dropped an extra line in there:
conclusions
(1) p
(2) r
(3) q
(4) p → q
(5) a is b
justifications
Given
Theorem 1 (modus ponens using (1))
Theorem 2 (modus ponens using (2))
Conditional introduction using (1) and (3)
Equivalence with (4)
#9 Consider
Axiom 1: s → t
Theorem 1: p → q
Theorem 2: q → ¬ t
Prove the following:
Theorem: If p, then s is impossible.
Proof
Given: p
Prove: ¬ s
conclusions
(1) p
(2) q
(3) ¬ t
(4) ¬ s
justifications
Given
Theorem 1 (modus ponens using (1))
Theorem 2 (modus ponens using (2))
Axiom 1 (modus tollens using (3))
#13 Suppose that is has been established that p implies either r or s, and
Theorem 1: y → ¬ p
Theorem 2: r → x
Theorem 3: s → y
Theorem 4: x → q
Prove, in outline form, that p → q.
Solution: The first thing to note is that “suppose it has been established” means “hey, this is
also a theorem of the system.” So, we have infact five theorems to work with (and nothing else).
Theorem
Theorem
Theorem
Theorem
Theorem
1:
2:
3:
4:
5:
y → ¬p
r→x
s→y
x→q
p → (r ∨ s)
Prove the following:
Theorem: If p, then q
Proof
Given: p
Prove: q
That sets up the structure. Now the other thing of interest is we have an “or” statement. An
“or” presents us with cases - we know either r or s, but not which one in particular. The trick
is to consider each in turn, and show that they both lead to the same conclusion. This is using
the rule of inference known as separation of cases.
conclusions
(1) p
(2) r ∨ s
(3) r
(4) x
(5) q
(6) r → q
(7) s
(8) y
(9) ¬ p
(10) p ∧ ¬ p →←
(11) ¬ s
(12) r
(13) q
justifications
Given
Theorem 5 (modus ponens using (1))
Assumption
Theorem 2 (modus ponens using (3))
Theorem 4 (modus ponens using (4))
Conditional introduction ((3)-(5))
Assumption
Theorem 3 (modus ponens using (7))
Theorem 1 (modus ponens using (8))
Conjunction of (1) and (9)
Assumption in (7) leads to contradiction
Disjunctive syllogism using (2) and (11)
Modus ponens using (6) and (12)
This one has a few other things of note. I started off treating this as a separation of cases assume r, see what happens, assume s, see what happens. In a true separation of cases, both
would have led to q. However, assuming s led to a contradiction ... which simply allowed me to
conclude NOT s, and backtrack a bit - combining ¬ s with r∨s through disjunctive syllogism simply means that r is no longer just an assumption, but an established result ... leading in turn to q.
Had I noticed in advance that s was going to blow up, I could have shortened the proof a bit by
considering it first:
conclusions
(1) p
(2) r ∨ s
(3) s
(4) y
(5) ¬ p
(6) p ∧ ¬ p →←
(7) ¬ s
(8) r
(9) x
(10) q
justifications
Given
Theorem 5 (modus ponens using (1))
Assumption
Theorem 3 (modus ponens using (3))
Theorem 1 (modus ponens using (4))
Conjunction of (1) and (5)
Assumption in (3) leads to contradiction
Disjunctive syllogism ((2) and (7))
Theorem 2 (modus ponens using (8))
Theorem 4 (modus ponens using (9))
The point here however is that at the outset, there was no particular reason to consider s first
- it was after seeing what happened in the first proof I noticed it could be done more efficiently - i.e., there’s nothing wrong with either proof, just assume both cases of the “or” and
see where they lead. If the both lead to the same result, you’ve done a separation of cases, and
if one leads to a contradiction, you’ve eliminated a case and get to keep the remaining one as fact.
#14 This one is done as a live one - there’s a lot of going in circles with the or ’s and the implications...
or, you can prove the contrapositive. Go watch.
Suppose the following have been established or assumed:
Axiom 1: p → ¬ y
Axiom 2: ¬ q → r
Theorem 1: p → ¬ z
Theorem 2: x → (q ∨ z)
Theorem 3: r → (x ∨ y)
Prove the following:
Theorem: If p, then q.
Proof
Given: p
Prove: q
conclusions
(1) p
(2) ¬ y
(3) ¬ z
(4) Now what?
justifications
Given
Axiom 1 (modus ponens using (1))
Theorem 1 (modus ponens using (2))