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consider the statement, if today is Monday then tomorrow is Tuesday

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how is the third condition true in this case?

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    This is a bad example of implication as "today is Monday if and only if tomorrow is Tuesday." Try "if it is raining, then the streets are wet." – abiessu Aug 21 '13 at 05:27
  • what is the question (i just don't see a question here) and what are you studying that lead you to ask this question , (is ist about intuitionistic logic of just classical logic) e ) – Willemien Aug 22 '13 at 07:32

2 Answers2

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Your understanding of the situation (that Tuesday only follows Monday) is more precise than the statement you have made.

You said "if today is Monday then tomorrow is Tuesday" which leaves open the possibility that tomorrow is Tuesday without today being Monday.

What you meant to say is "if and only if (iff) today is Monday then tomorrow is Tuesday". This would be represented by the equivalence operator (⇔), not the implication operator (⇒), and would produce the truth table that you expect (TFFT instead of TFTT)

Sparr
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By the definition of the logical connective $\to$, $p\to q$ is defined to be true for every combination of truth values of $p$ and $q$ except $p$ being true and $q$ false.

This isn’t quite like normal English usage of if $p$ then $q$, which generally assumes some genuine connection between $p$ and $q$. The formal logical connective is defined without regard to any possible meaning of $p$ and $q$: its truth value is simply a function of the truth values of $p$ and $q$.

However, some intuitive justification of the definition is possible. What evidence would you require in order to be certain that an implication $p\to q$ was false? The only absolutely certain evidence would be a demonstration that $p$ is true and $q$ is not. Thus, we say in that case that $p\to q$ is false, and since we must (in classical logic) assign to each of the other three possibilities either the value True or the value False, we give truth the benefit of the doubt when $p$ is false.

However, your specific example muddies the issue because for the particular $p$ and $q$ that you’ve chosen, it is not possible for one to be true and the other to be false: it’s actually the case that $p\leftrightarrow q$, and only the top and bottom lines of your table are actually possible states of the universe.

Brian M. Scott
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  • Your final argument implies that formulating the intuitive law of commutation that "antecedents of a conditional with more than one antecedent commute" as CCpCqrCqCpr muddies two-valued proposition calculus, since if CpCqr is false, then p=1, q=1, and r=0. So, CqCpr is also false. Thus, your final argument would require formulating the principle that the antecedents of a conditional commute as an equivalence (or at least a much more complicated formula than CCpCqrCqCpr in a C-N language). – Doug Spoonwood Aug 21 '13 at 16:48