Incomplete question:
We can take this example:
Construct a truth table for each of these compound propositions.
a) p ∧ ¬p
b) p ∨ ¬p
c) (p ∨ ¬q) → q
d) (p ∨ q) → (p ∧ q)
e) (p → q) ↔ (¬q → ¬p)
f) (p → q) → (q → p)
Solution:
Given:
a) p∧¬p
b) p∨¬p
Answer:
p01¬p10p∧¬p00p∨¬p11
c) (p∨¬q)→q
d) (p∨q)→(p∧q)
e) (p→q)↔(¬q→¬p)
f) (p→q)→(q→p)
Answer:
p0011q0101¬p1100¬q1010p∨¬q1011(p∨¬q)→q0101p∨q0111p∧q0001(p∨q)→(p∧q)1001p→q1101¬q→¬p1101q→p1011(p→q)↔(¬q→¬p)1111(p→q)→(q→p)1011
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