SE 500 (Math for SE)   Fall 2023
HW #3: Proofs involving Equivalence, Negation, and Inequivalence
Due: 6:30pm, Tuesday, Sept. 18

1. In each step in the proof below, Leibniz (sometimes with an assist from Substitution) is applied. That is, in going from each line to the next, some subexpression E is replaced by an expression F such that E = F (or possibly F = E) is an instantiation of (i.e., the conclusion of applying Substitution to) one of the theorems in our arsenal. (That is, E = F (or else F = E) is P[r:=Q], where P is a theorem in our arsenal, r is a list of variables, and Q is a list (of length equal to r) of expressions.)

Here, the only theorems used are axioms (3.1), (3.2), and (3.3). Fill in each "hint" so as to inform the reader which theorem (and which instantiation of it) was used. The first hint is filled in for you. Also, somehow highlight the subexpressions E and F in each step (using underlining or circling or something). In the first step below, the former is underlined and the latter is in italics. 

   (r ≡ q) ≡ (true ≡ (q ≡ p))

=      < (3.3)[q:=r] >

   (r ≡ q) ≡ ((r ≡ r) ≡ (q ≡ p))

=      <                                                 >

   (r ≡ q) ≡ (r ≡ (r ≡ (q ≡ p)))

=      <                                                 >

   (r ≡ q) ≡ (r ≡ ((r ≡ q) ≡ p))

=      <                                                 >

   (r ≡ q) ≡ (((r ≡ q) ≡ p) ≡ r)

=      <                                                 >

   (r ≡ q) ≡ ((r ≡ q) ≡ (p ≡ r))

=      <                                                 >

   ((r ≡ q) ≡ (r ≡ q)) ≡ (p ≡ r)

=      <                                                 >

   true ≡ (p ≡ r)

=      <                                                 >

   p ≡ r

2. On page 47, Gries proves theorem (3.11) (¬p ≡ q ≡ p ≡ ¬q) by transforming it to (and thereby showing it to be equivalent to) an instance of theorem (3.5) (p ≡ p). Here, you are asked to prove (3.11) in two different ways:

(a) Interpreting (3.11) to say (¬p ≡ q) ≡ (p ≡ ¬q), prove it by transforming its left-hand side ¬p ≡ q into its right-hand side p ≡ ¬q.

(b) Interpreting (3.11) to say ¬p ≡ (q ≡ p ≡ ¬q), prove it by transforming its right-hand side q ≡ p ≡ ¬q into its right-hand side ¬p.

3. Do Exercise 3.10, which is to prove    (3.14)   (p ≠ q) = ¬p ≡ q

(Note that ≠ is used here rather than ≡ with a line through it. Why? Because the instructor could not find how to produce the latter symbol using HTML.)

4. Do Exercise 3.12, which is to prove the associativity of ≠:

(3.17) ((p ≠ q) ≠ r) ≡ (p ≠ (q ≠ r))

using the heuristic of Definition Elimination (3.23).

5. Do Exercise 3.14, which is to prove mutual interchangability of ≡ and ≠ (3.19):

(3.19)   p ≠ q ≡ r   ≡   p ≡ q ≠ r

using the heuristic of Definition elimination(3.23).

6. Prove the theorem ((p ≡ q) ≡ (false ≡ (q ≡ r))) ≡ (r ≠ p)

Note that, due to the associativity of ≡ (Axiom (3.1)) and the mutual associativity of ≡ and ≠ (Theorem (3.18)), this theorem can be written without any parentheses at all.