Question

Using a Truth table, determine the value of the compound proposition(10marks)

((𝑝 ∨ 𝑞) ∧ (￢𝑝 ∨ 𝑟)) → (𝑞 ∨ 𝑟).

Answer 1

$p~~~q~~~r~~~p\lor q~~~\lnot p\lor r~~~(p\lor q)\land(\lnot p\lor r)\\ 0~~~0~~~0~~~~~~0~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~0\\ 0~~~0~~~1~~~~~~0~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~0\\ 0~~~1~~~0~~~~~~1~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~1\\ 0~~~1~~~1~~~~~~1~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~1\\ 1~~~0~~~0~~~~~~1~~~~~~~~~~~~0~~~~~~~~~~~~~~~~~~~0\\ 1~~~0~~~1~~~~~~1~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~1\\ 1~~~1~~~0~~~~~~1~~~~~~~~~~~~0~~~~~~~~~~~~~~~~~~~0\\ 1~~~1~~~1~~~~~~1~~~~~~~~~~~~1~~~~~~~~~~~~~~~~~~~1\\$

$Let\space s=(p\lor q)\land(\lnot p\lor r)$

$p~~~q~~~r~~~s~~~q\lor r~~~s\rarr(q\lor r)\\ 0~~~0~~~0~~~0~~~~~~0~~~~~~~~~~~~~~1\\ 0~~~0~~~1~~~0~~~~~~1~~~~~~~~~~~~~~1\\ 0~~~1~~~0~~~1~~~~~~1~~~~~~~~~~~~~~1\\ 0~~~1~~~1~~~1~~~~~~1~~~~~~~~~~~~~~1\\ 1~~~0~~~0~~~0~~~~~~0~~~~~~~~~~~~~~1\\ 1~~~0~~~1~~~1~~~~~~1~~~~~~~~~~~~~~1\\ 1~~~1~~~0~~~0~~~~~~1~~~~~~~~~~~~~~1\\ 1~~~1~~~1~~~1~~~~~~1~~~~~~~~~~~~~~1\\$

((𝑝 ∨ 𝑞) ∧ (￢𝑝 ∨ 𝑟)) → (𝑞 ∨ 𝑟) is always true.