JUAN CARLOS Q. GARCIA
BSIT 202A
1) Show, by the use of the truth table/matrix, that the statement (p v q) ∧ [( ¬p) ∧ (¬q)] is a
tautology.
p
q
T
T
F
F
T
F
T
F
¬p
F
F
T
T
¬q
F
T
F
T
pvq
T
T
T
F
(¬P)^(¬q)
F
F
F
T
(pvq)^[(p ¬)^(vq)]
F
F
F
F
This statement is not tautology, it is contradiction because this formula is always
false.
2) Show that p↔ q and ( p ^ q) ∨ (¬p^¬q) are logically equivalent.
Ang una nating gagawin
Gawin nating truth table ang parehas na statement, lets denote show
that p↔ q as S1, ( p ^ q) as A (¬p^¬q) as B, ( p ^ q) ∨ (¬p^¬q) as S2.
p
q
A
B
S1
S2
T
T
T
F
T
T
T
F
F
F
F
F
F
T
F
F
F
F
F
F
F
T
T
T
Pangalawang method na gagawin
p↔q=(p → q)∧(q → p)=(¬p ∨ q)∧(¬q ∨ p)=
=(¬p∧(¬q ∨ p))∨(q∧(¬q ∨ p))=((¬p∧¬q)∨(¬p ∧ p))∨((q∧¬q)∨(p ∧ q))
(¬p ∧ p) ay always falls, as well as (q∧¬q), since (x∨ False) is true only if x is true, (x∨ False)=x.
Using this we can remove (¬p ∧ p) and (¬q ∧ q), which gives as
P ↔ q=(p ∧ q)∨(¬p∧¬q)