Use the truth tables method
Question
- Use the truth tables method to determine whether (¬p v q)∧(q → ¬r ∧ ¬p)∧ (p v r) is satisfiable.
- Is the compound proposition ¬(P v Q v R) a contradiction? Justify.
- Prove using the laws of logic that A → (B v C0=(A → B) v( A → C)
- Prove using the laws of logic that (¬Q → ¬p) → (p → Q) = true
- Next prove using the laws of logic that (p → r) ∧ (q → r) = ((p v q) )→ r
Summary
Here we have given 5 questions. So We have to use the truth tables method to determine the given condition. And Next, we have to answer if a given condition is a contradiction. So that in the third, fourth, and fifth questions we have to prove the logic of the given condition.
Explanation
So here in the first question, we are using a truth table. In that truth table, a proposition is said to be satisfied if anyone one of its output/results is true.
| P | Q | R | ¬P | ¬R | ¬PvQ | Q → R | Q→ R ∧ ¬P | (→ PVQ) ∧ Q→ R | PVR | Result |
| T | T | T | F | F | T | F | F | F | T | F |
| T | T | F | F | T | T | T | F | F | T | F |
| T | F | T | F | F | F | T | F | F | T | F |
| T | F | F | F | T | T | T | F | F | T | F |
| F | T | T | T | F | T | F | F | F | T | F |
| F | T | F | T | T | T | T | T | T | F | F |
| F | F | T | T | F | F | T | T | F | T | F |
| F | F | F | T | T | T | T | T | T | T | F |
The given propositions are not a satisfaction as all the results are false.
For the given proposition contradiction is false. And it can be driven out of it. So, in the given condition
| P | Q | R | PV Q | Pv Q v R | (P v Q v R) |
| T | T | T | T | T | F |
| T | T | F | T | T | F |
| T | F | T | T | T | F |
| T | F | F | T | T | F |
| F | T | T | T | T | F |
| F | T | F | T | T | F |
| F | F | T | F | T | F |
| F | F | F | F | F | T |
Here out of eight, seven outputs are false. As the majority is on the side of false. It can be said that the given condition is a contradiction.
We have to prove the given expression.
A→(B v C)
so that we can say that
A → (B or C)
A→B or A→C
(A→B) V (A→C)
Next, we have to prove (¬Q → ¬p) → (p → Q) = true
(– Q → –P) → (P → Q)
–(Q → P) → (P → Q)
(P → Q) → ( P → Q)
=true
Hence proved
Here we have to prove (p → r) ∧ (q → r) = ((p v q) )→ r
(p → r) ∧ (q → r) So, here,
p implies to r and q implies to r
So here both p and q implies to r
(p v q) → (r)
any one of p and q can imply to r
Also read, given pair of 16-bit binary data