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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))