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Give the possible truth values of

  • a) $P \Rightarrow (Q \Rightarrow R)$
  • b) $(P \Rightarrow Q) \Rightarrow R$

Match?

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    Try with truth-table. – 2017-01-28

2 Answers 2

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a.) P ⇒(Q⇒R) ⇐⇒ (P ⇒Q)⇒(P ⇒R)

So taking the right hand side of expression create truth table.

Similarly for option b.) simplify expression and create truth table.

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    Sorry, but the conditional is *not* associative! – 2017-01-28
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    The first equivalence is valid, but I don't see how it is _useful_ for this question. – 2017-01-28
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    @Bram yes I see it. In some cases both are not equivalent. – 2017-01-28
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    @Henning Makholm its help in creating truth table. – 2017-01-28
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    How is rewriting to a _longer, more complicated_ formula helpful for filling out a truth table? – 2017-01-28
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    I don't think its complicated. – 2017-01-28
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    But I will try more easy solution wait. – 2017-01-28
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    My answer is wrong so you can unaceept it. – 2017-01-29
1

The conditional is not associative, so these are not the same. Try a truth table, or just see what happens if $P$ and $R$ are both False.

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    @l530332 please note the answer you accepted is not correct ... – 2017-01-28