I have gone through various sites but i can't understand the justification for the truth table of "If $p$ then $q$". Is it accepted by the mathematicians without any proof or justification? \begin{array}{c|c|c} p & q & p\rightarrow q \\ \hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \end{array} Last two values of truth table seems a bit confusing how is true?
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You can find many many similar questions on this site, like: truth-table-fo $p \to q$ and understanding-implication-in-logic-truth-tables – Mauro ALLEGRANZA Dec 13 '17 at 10:12
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that last two values seems confusing. i have read that article too........ Suppose p=it is cloudy outside,q=it is raining outside then according to p->q if it is cloudy outside then it is raining outside how does this statement justify this – Aniket Thomas Dec 13 '17 at 10:14
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there is no justification; it is the definition of "if.... then....". – the4seasons Dec 13 '17 at 10:20
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See also the post: implications and ordinary language about a "reasonable" justification. – Mauro ALLEGRANZA Dec 13 '17 at 10:21
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why there is no justification for this? – Aniket Thomas Dec 13 '17 at 10:21
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See also the post: conditional statements "only if". – Mauro ALLEGRANZA Dec 13 '17 at 10:22
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See also the post: on the truth value of implication connective. – Mauro ALLEGRANZA Dec 13 '17 at 10:23
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Stupid example: For integers $a,b,c$ we sure want $((a=b) \Longrightarrow (ac=bc))$ to be true. For $a\neq b$ and $c\in{0,1}$ this justifies the last two rows of your truth table. – Christoph Dec 13 '17 at 10:24
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See also the post: how to interpret material conditional and explain it to freshmen. – Mauro ALLEGRANZA Dec 13 '17 at 10:31
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See also the post: in classical logic why is $p \rightarrow q$ true if p is false and q is true. – Mauro ALLEGRANZA Dec 13 '17 at 10:33
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@MauroALLEGRANZA Can we have more? – Christoph Dec 13 '17 at 10:35
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Thanks @MauroAllegranza for everything – Aniket Thomas Dec 13 '17 at 11:05
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I know this table as the definition of the implication. – Dec 13 '17 at 11:43
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See also: Logical Conditional Rationale – Bram28 Dec 13 '17 at 17:07
4 Answers
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There are 16 possible truth tables for two binary variables $p$ and $q$ that take the values $T$ or $F$.
Assuming that we want:
I. $p = T$, $q = T$ and $p \implies q = T$
II. $p = T$, $q = F$ and $p \implies q = F$
we have that from the 16 possible truth tables only these 4 comply with conditions I and II:
\begin{array}{|c|c|c|c|c|c|} p & q & 1 & 2 & 3 & 4\\ \hline T & T & T & T & T & T\\ T & F & F & F & F & F\\ F & T & F & T & F & T\\ F & F & F & F & T & T\\ \end{array}
The truth table $1$ corresponds to the truth table of $p \land q$ so we discard that this table is suitable for $p \implies q$.
The truth table $2$ corresponds to $q$ so we discard that this table is suitable for $p \implies q$.
The truth table $3$ corresponds to the truth table of $p \Longleftrightarrow q$ so we discard that this table is suitable for $p \implies q$.
Therefore table $4$ is the only truth table suitable for $p \implies q$ if we think that conditions I and II must be fulfill and $p$ and $q$ are two binary variables that take the values $T$ or $F$.
That is the justification for the truth table of $p \implies q$.
luifrancgom
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For the third when $p$ is false prove $p\to q$ by contradiction.
So accept $p$ and assume $\neg q$.
As $\neg p$ is true, $p$ and $\neg p$ are a contradiction.
Thus the assumption $\neg q$ is false.
This proves $p \to q$ when p is false.
The fourth is simular.
Christoph
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William Elliot
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You can memorise easily it: $$a \rightarrow b = \overline a \lor b$$
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This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review – José Carlos Santos Dec 15 '17 at 00:00
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@JoséCarlosSantos This rule helps to understand why truth table of this function is that. – Devid Otladka Dec 15 '17 at 00:09
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$$ \left( p \implies q \right) \equiv \mbox{“p implies q”} \equiv \mbox{“if p then q”} $$
So if the statement "$p \implies q$" holds, we have:
If $p$ is true then the statement tells us $q$ must be true.
If $p$ is false then the statement does not tell us anything about $q$.
The only way to prove the statement "$p \implies q$" is wrong is to show that $p$ holds but $q$ does not. That is why the truth table only indicates the statement is "false/wrong" in that one row.
Michael
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