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 |
(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.
(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.)
using the heuristic of Definition Elimination (3.23).
using the heuristic of Definition elimination(3.23).
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.