Logic with Elements of Set Theory
Truth Tables
October 2013
0. Assign truth values (T or F ) to the following statements:
4. The equation x2 + 1 = −5 has a real solution or five is an odd integer divisible by 3.
1. 3 6 4 and 4 is an odd integer.
2. 3 6 4 or 4 is an odd integer.
5. 21 · 10666 − 1 is divisible by 6 or 2x + 3 = 7.
3. 5 is odd or divisible by 4.
A proposition can be
• valid (a tautology), when it is always true, for all possible substitutions of variables, like
(−p ∧ −q) ⇔ −(p ∨ q).
• satisfiable (a contingency), when it holds true for at least one substitution, like p ⇒ q.
• a contradiction, when it is never true, like p ∧ −p.
1. For aech of the following construct a truth table and determine whether the
proposition is valid, satisfiable or a contradiction.
21. p ⇒ (q ⇒ (q ⇒ p)),
1. p ⇒ (p ∧ q),
11. (p ∧ q) ⇒ (p Y q),
2. −p ∨ (p ∨ q),
12. (p Y q) Y (p Y −q),
3. (p ∨ −q) ⇒ q,
13. (p ∨ q) Y (p ∨ −q),
4. (p ⇒ q) ⇒ q,
14. (p ⇒ q) Y (q ⇒ −p),
5. p ⇒ (q ⇒ q),
15. (p ⇒ q) Y (q ⇒ p),
24. ((p ∨ q) ∧ −q) ⇒ p,
6. (−p ⇒ q) ∨ p,
16. (p ⇒ −q) Y (q ⇒ p),
25. ((p ⇒ q) ∧ −q) ⇒ −p,
7. p ∧ (−p ∧ q),
17. (p ⇒ q) Y (−q ⇒ p),
8. p ∧ −(p ∨ −q),
18. (p ∧ −q) ⇒ (−r ⇒ p),
9. p ∧ −(p Y −q),
19. (−p ∨ q) ⇒ (p ⇒ q),
10. (p ∧ q) ⇒ (p ∨ r),
20. (p ⇒ q) ⇒ (q ⇒ p),
22. (p Y q) ∨ (p ∧ q),
23. (p ∧ q) ⇒ (−p ∨ −q),
26. (p ⇒ (q ∧ r)) ⇒ (p ⇒ r),
27. (p ∧ (p ⇒ q)) ⇔ (p ∧ −q),
28. ((p ∧ q) ∨ −(p ∨ q)) ⇒ (p ⇔ q).
2. For each of the following give an example of a proposition which is false only in
the given case.
1. p, q and r are false,
4. q and r have the same truth value, different to that of p,
2. p, q and r are true,
3. p and q have the same truth value,
5. p and q are true and r is false.
3. Which propositions from 1 are logically equivalent?
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Examples of solutions.
Ex. 0.5 It is not possible to decide, for 2x + 3 = 7 is neither true nor false. For different x’s we
have different results.
Ex. 1.6.
(−p ⇒ q) ∨ p.
p
0
0
1
1
q (−p ⇒ q)
0 1
0
1 1
1
0 0
1
1 0
1
∨ p
0
1
.
1
1
The proposition is satisfiable. It is equivalent to the alternative p ∨ q—the True/False distribution
is the same.
(p ⇒ q) Y (q ⇒ −p).
Ex. 1.14
q (p ⇒ q)
0
1
1
1
0
0
1
1
p
0
0
1
1
Y
0
0
1
1
(q ⇒ −p)
1
1
1
1
1
0
0
0
The proposition is satisfiable and obviously equivalent to p.
(p ⇒ q) ⇒ (q ⇒ p).
Ex. 1.20
p
0
0
1
1
q (p ⇒ q)
0
1
1
1
0
0
1
1
⇒
1
0
1
1
(q ⇒ p)
1
0
1
1
The proposition is satisfiable and obviously equivalent to q ⇒ p.
Ex. 1.21
p ⇒ (q ⇒ (q ⇒ p)).
p
0
0
1
1
q
0
1
0
1
p⇒
1
1
1
1
(q ⇒ (q ⇒ p))
1
1
0
0
1
1
1
1
The proposition is a tautology.
Ex. 2.5
“(p ∧ q) ⇒ r” is false only when p and q are true and r is false.
Further reading.
http://college.cengage.com/mathematics/larson/intermediate_algebra_gf/3e/shared/appendices/
2263app_e1.pdf
http://www.math.umn.edu/~jodeit/course/ACaRA01.pdf
http://www.cl.cam.ac.uk/teaching/1112/DiscMathI/exercises.pdf
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