Unit 3 - Linear Relationships
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Discrete Mathémátics nome > 1.5 : Laws of propositional logic Nikolau Help/FAQ Need help with this 608790.4733550.qx3zqy7 Jump to level 1 Simplify neg (qto neg neg q) to F 1. Select a law from the right to apply Laws neg (qto neg neg q) Distributive Complement (a∧b)v(a∧c) a∧(bvc) asqrt(^a) (avb)∧(avc) av(b∧c) Commutative bva awedge b bwedge dentity De Morgan's a^b avee neg b v ≡ a ¬(avb) wedge neg b Double negation Conditional a ato bE | neg avee b arightarrow b= 2 3
Gauth ai solution, super gauth ai, explanation.
To simplify the expression \neg (q \to \neg \neg q) ¬ ( q → ¬¬ q ) , we start by recalling the definition of implication in propositional logic. The expression q \to \neg \neg q q → ¬¬ q can be rewritten using the equivalence p \to q \equiv \neg p \vee q p → q ≡ ¬ p ∨ q . Thus, we have:
q \to \neg \neg q \equiv \neg q \vee \neg \neg q q → ¬¬ q ≡ ¬ q ∨ ¬¬ q
Since \neg \neg q ¬¬ q simplifies to q q , we can rewrite the expression as:
\neg (q \to \neg \neg q) \equiv \neg (\neg q \vee q) ¬ ( q → ¬¬ q ) ≡ ¬ ( ¬ q ∨ q )
Next, we apply De Morgan's Law, which states that \neg (A \vee B) \equiv \neg A \wedge \neg B ¬ ( A ∨ B ) ≡ ¬ A ∧ ¬ B . Therefore, we have:
\neg (\neg q \vee q) \equiv \neg \neg q \wedge \neg q ¬ ( ¬ q ∨ q ) ≡ ¬¬ q ∧ ¬ q
This simplifies to:
q \wedge \neg q q ∧ ¬ q
The expression q \wedge \neg q q ∧ ¬ q is a contradiction, which is always false. Therefore, we conclude that:
\neg (q \to \neg \neg q) \equiv F ¬ ( q → ¬¬ q ) ≡ F
Thus, the final answer is:
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