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Mathematics · Ch 1 — Mathematical Logic

Some important results

1.4.1

Some important results

Four equivalences underpin most of the simplification work in this chapter. This section proves the first two with a single truth table and leaves the other two as a truth-table activity.

i) p→q≡ p∨qp → q ≡ ~p ∨ q

ii) p↔q≡(p→q)∧(q→p)p ↔ q ≡ (p → q) ∧ (q → p)

iii) p∨(q∧r)≡(p∨q)∧(p∨r)p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

iv) p∧(q∨r)≡(p∧q)∨(p∧r)p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)

Proving (i) and (ii).

pq~pp→qq→pp↔q~p∨q(p→q)∧(q→p)
TTFTTTTT
TFFFTFFF
FTTTFFTF
FFTTTTTT

The p→q column and the ~p∨q column are identical (T,F,T,T both), proving p→q≡ p∨qp → q ≡ ~p ∨ q. The p↔q column and the (p→q)∧(q→p) column are also identical (T,F,F,T both), proving p↔q≡(p→q)∧(q→p)p ↔ q ≡ (p → q) ∧ (q → p) — confirming that a biconditional is nothing more than 'the implication holds in both directions'.

Activity — proving (iii) and (iv) by truth table. With three variables there are eight rows to check:

pqrp∨(q∧r)(p∨q)∧(p∨r)p∧(q∨r)(p∧q)∨(p∧r)
TTTTTTT
TTFTTTT
TFTTTTT
TFFTTFF
FTTTTFF
FTFFFFF
FFTFFFF
FFFFFFF
Table 1Table 1.20 — Proof that p → q ≡ ~p ∨ q and p ↔ q ≡ (p → q) ∧ (q → p)
pq~pp → qq → pp ↔ q~p ∨ q(p → q) ∧ (q → p)
TTFTTTTT
TFFFTFFF