Classify the following as tautologies, contradictions or contingencies using logical equivalences.

Can anyone let me know what I'm missing or doing wrong? I got stuck, here is what I have so far:

$\lnot(\lnot x \rightarrow (y \rightarrow z)) \rightarrow (y \rightarrow(x \lor z))$

$≡ \lnot(\lnot x\rightarrow (y \rightarrow z)) \rightarrow (\lnot y \lor(x \lor z))$ Implication Equivalence

$ \equiv \lnot(\lnot x \rightarrow (\lnot y \lor z)) \rightarrow (\lnot y \lor (x \lor z))$ // Implication Equivalence

$≡ ¬(\lnot\lnot x\lor (\lnot y \lor z)) \rightarrow (\lnot y \lor x \lor z)$ Implication Equivalence, Associativity

$≡ \lnot\lnot (\lnot \lnot x \lor \lnot y \lor z) \lor (\lnot y \lor x \lor z)$ Implication Equivalence, Associativity

$≡ (x \lor \lnot y \lor z) \lor (\lnot y x \lor z) $ Double Negation, twice used.

$ ≡ (x \lor x) \lor (\lnot y \lor \lnot y) \lor (z \lor z)$ Associative and Commutative Law

$≡ x \lor \lnot y \lor z$ // Idempotent law

I just don't know what rules to use or what to do from here.



Your work is fine. You have not done anything wrong in your work. You just stopped short of determining if the proposition you end with (and hence the original proposition) is a tautology, contradiction, or contingency.

Now we need to interpret the result, with the last line logically equivalent to the original proposition, we can look at the last line, $x \lor \lnot y \lor z$, to evaluate if and when it is true/false. This can be done easily by using a truth-table (you'll find one row evaluates to false, the rest true).

The proposition is false when $x, z$ are both false and $y$ is true. (Can you see that $x \lor \lnot y \lor z$ would be false under that truth value assignment?)

Otherwise, it is true. That is, it is true if $x$ is true (independently of the truth value assignment of $y$ or $z$.) Likewise, it is true if $y$ is false, and/or it is true when $z$ is true.

Since the truth-value of the proposition depends on (is contingent upon) the truth-value assignments of the variables, the proposition is a contingency.

  • 1
    $\begingroup$ Thank you so much for clearing that up, I was too busy trying out laws and rules instead of stepping away and actually using logic haha $\endgroup$
    – Karim B
    Jan 22 '15 at 23:26
  • $\begingroup$ You're welcome, Karim! $\endgroup$
    – amWhy
    Jan 23 '15 at 12:42

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