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((~p ∧ q) ∧ (q ∧ r)) ∧ ~q a tautology, a contradiction, or neither a tautology nor a contradiction?

User ChapMic
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1 Answer

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Final Answer:

((~p ∧ q) ∧ (q ∧ r)) ∧ ~q is neither a tautology nor a contradiction. It is a contingency.

Step-by-step explanation:

Let's break down the proposition step-by-step:

(~p ∧ q): This means "not p and q." If q is true, then this conjunction is true regardless of the value of p. If q is false, then the conjunction is true only if p is true.

(q ∧ r): This means "q and r." If q is false, then this conjunction is false regardless of the value of r. If q is true, then the conjunction is true only if r is true.

((~p ∧ q) ∧ (q ∧ r)) ∧ ~q: This combines the previous two conjunctions with "and not q." Since we already established that the conjunction is true only if q is true, and now we negate q, the entire proposition becomes false.

However, this is not true for all possible values of the variables. Consider the following truth table:

p q r (~p ∧ q) (q ∧ r) ((~p ∧ q) ∧ (q ∧ r)) ∧ ~q

T T T F T F

T T F F F F

T F T F F F

T F F F F F

F T T T T F

F T F T F F

F F T T F F

F F F T F F

As you can see, the proposition is false in all cases except one: when p is true, q is true, and r is true. Therefore, it is not a tautology (always true) or a contradiction (always false). It is a contingency, meaning its truth value depends on the specific assignment of truth values to the variables.

User Manmeet
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