Discrete Mathematics Homework
1.3.54
(a)
p p↓p ¬p
T F F
F T T
Therefore, p↓p is logically equivalent to ¬p
(b)
p q p↓q (p↓q)↓(p↓q
)
p∨q
T T F T T
T F F T T
F T F T T
F F T F F
Therefore,(p↓q)↓(p↓ q) is logically equivalent to p∨q
(c)Because {¬,∨} is a functionally complete collection of logical
operators, and we have: ¬p≡p↓p and p∨q≡(p↓q)↓(p↓ q), so {↓} is a
functionally complete collection of logical operators.
1.3.34
We know (1)a→b≡¬a∨b, (2)¬(a∧b)≡¬a∨¬b, (3)¬(a∨b)≡¬a∧¬b,
(4)a∨(¬b∧¬a) ≡a∨¬b, (5)a∨¬a≡a, (6)a∨T≡T
So (p∨q)∧(¬p∨r)→(q∨r)
≡¬((p∨q)∧(¬p∨r))∨(q∨r) (use(1))
≡(¬(p∨q)∨¬(¬p∨r))∨(q∨r) (use(2))
≡((¬p∧¬q)∨(p∧¬r))∨(q∨r) (use(3))
≡(¬p∧¬q)∨(p∧¬r)∨q∨r
≡(q∨(¬p∧¬q))∨(r∨(p∧¬r))
≡(q∨¬p)∨(r∨p) (use(4))
≡q∨r∨(p∨¬p)
≡q∨r∨T (use(5))
≡T (use(6))
Therefore, this is a tautology.
1.3.54
(a)
p p↓p ¬p
T F F
