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Many theorems in math have an if-then form. For example: "If a polynomial is of $n^{th}$ degree, then it has $n$ roots. In my other question, I learned that in order to analyze statements using truth tables, the statements must be completely independent. However, I'm not sure that anything in math is independent. Everything can be proven from the axioms as far as I know, so if $p \implies q$, the truth value of $p$ automatically determines the truth value of $q$. So if the statement "$f$ is polynomial is a degree $5$" automatically makes the statement $f$ has $5$ roots", the truth table seems nonsensical. Do mathematical if-then statements have anything to do with the classical if-then statements from logic?

$$p \implies q$$

$$\begin{array}{|c|c|c|} \hline p&q&p\implies q\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T\\\hline \end{array}$$

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Ovi
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    Why do you think that the truth table is non-sensical ? It shows false, if $p$ is true AND $q$ is false. In the other cases, it shows true. What should be non-sensical ? – Peter Jun 04 '16 at 18:12
  • Additionally, there are true, but unprovable statements in mathematics. – Peter Jun 04 '16 at 18:14
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    yes , if a then b else c; is translated to a*b+not(a)*c – Abr001am Jun 04 '16 at 18:27
  • @Peter Well from the answers of the question linked in the OP, I understood that my truth table was nonsensical because $p$ and $q$ were not independent. $p \implies q$ is true when $p=F$, $q=T$, but in the example that I gave, $(F \implies T)=T$ was nonsensical, and the reason was that $p$ and $q$ were not independent. – Ovi Jun 04 '16 at 18:30
  • @Agawa001 Does "*" mean "$\implies$" and the $+$ mean "or"? – Ovi Jun 04 '16 at 18:34
  • The case $p$ false and $q$ true is in fact somewhat strange. But mathematically, it is correct to say , for example : If 2+2=5, then 2+2=4, which is especially weird. – Peter Jun 04 '16 at 18:35
  • @Peter I'm not sure if you took a look at the linked question, but I specified $p: x > 0$ and $q:$ The equation $100 = \sqrt x$ has a solution in $\mathbb{R}$ . Now if $p$ is false that implies $x \le 0$. We have $(F \implies T)=T$ logically, bu mathematically with those given statements we would've had $x \le 0 \implies$ The equation $100 = \sqrt x$ has a solution in $\mathbb{R}$, which is nonsensical mathematically. The resolution (according to those answers at least) is that $p$ and $q$ are not independent. – Ovi Jun 04 '16 at 18:43
  • @Ovi , obviously !!! – Abr001am Jun 04 '16 at 18:46
  • The point is : The statement : "If $x>0$ , then $100=\sqrt{x}$ has a solution" is true even in the case $x\le 0$, because the statement does not tell anything about the case $x\le 0$. – Peter Jun 04 '16 at 18:46
  • Moreover : "if $x>0$" is an unlucky formulation because such a statement normally implies that we have a fixed $x$. – Peter Jun 04 '16 at 18:49
  • Your statement should be formulated as follows : There is an $x>0$ with $100=\sqrt{x}$. In this case, you avoid any issues. – Peter Jun 04 '16 at 18:58
  • By the way, I remember that I wanted a resolution of Curry's paradox (If this sentence is true, then Santa Claus exists - or any other false statement ). Someone claimed that there are many resolutions but noone gave me one. – Peter Jun 04 '16 at 19:18
  • @Peter Well if you would post on any other question the statement $If x \le 0$, then the equation $100 = \sqrt x$ has a solution in $\mathbb{R}$ you would get 5 downvotes haha. There is something that just doesn't click with me with propositional logic; I'm not sure what it is, but I think part of it is this apparent "difference" in the treatment of if-then statements in logic and in math. I'm sure there isn't actually a difference but it just seems that way to me. – Ovi Jun 04 '16 at 20:29
  • Typically, mathematics deals with statements *not* of the form $P{\implies}Q,$ but of the form $\forall x (P(x){\implies}Q(x)),$ which truth tables generally can not handle. – ryang Apr 20 '23 at 11:05

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Do mathematical if-then statements have anything to do with the classical if-then statements from logic?

Yes; they use "the same" if-then.

Everything can be proven from the axioms

or, more precisely, in a mathematical theory every theorem is proved from the axioms.

This amount to:

if the axioms are true, then also the theorems are.

What happens if the axioms are not true? Well, the conditional still holds, but we are losing our time with a "wrong" mathematical theory.

A well-know quote from Bertrand Russell, The Principles of Mathematics page 3:

  1. Pure Mathematics is the class of all propositions of the form “$p$ implies $q$”, where $p$ and $q$ are propositions containing one or more variables, the same in the two propositions, and neither $p$ nor $q$ contains any constants except logical constants.

Examples of arithmetical theorems that are conditionals with false antecedent.

1) "if $2$ is odd, then $2=1$".

Proof

We define : $Even(n) := \exists z (n = z \times SS0)$.

Finally, we define: $Odd(x) := \lnot Even(x)$.

From Peano axioms it is easily proved that: $Even(SS0)$, i.e. $Even(2)$.

Thus, having proved $\lnot Odd(2)$, using the tautology:

$\lnot A \to (A \to B)$,

with $Odd(2)$ in place of $A$ and $2=1$ in place of $B$, by modus ponens we conclude with:

$Odd(2) \to 2=1$.

2) "if $2$ is odd, then $2=2$".

Proof

From equality axiom: $\forall x (x=x)$, we get: $2=2$.

Thus, using the tautology:

$A \to (B \to A)$,

with $2=2$ in place of $A$ and $Odd(2)$ in place of $B$, by modus ponens we conclude with:

$Odd(2) \to 2=2$.

Mauro ALLEGRANZA
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  • So my understanding is that when we are given a theorem in math as an $p \implies q$ statement, we are being told "This $p \implies w$ statement is always true, meaning that we can possibly find every combination of $p$ and $q$ except for $p=F$ and $q=T$"? I guess what I'm not really sure I get why my example here (http://math.stackexchange.com/questions/1793713/question-about-conditional-statements-as-applied-to-math) does not work (would really appreciate it if you could take a look). As an answer to that question, people said $p$ and $q$ were not independent. But what does that mean? – Ovi Jun 04 '16 at 20:43
  • ...Does it mean that it's not valid to put $p$ and $q$ in a $p \implies q$ relationship? Does it mean it's to put in in $p \implies q$ as long as $q$ as long as $q$ is true? – Ovi Jun 04 '16 at 20:44
  • @Ovi - in math usually we are not interested into conditional whatever like "if $0=1$, then the moon is round", but to theorems proved from axioms. Thus, the conditional we are using are like: "if axiom $A$ holds, then theorem $T$ holds also". The fact that the conditional is true also when $A$ is false is a fact than we accept but of course is of no relevance for the mathematical theory we are workin whit. – Mauro ALLEGRANZA Jun 05 '16 at 07:36
  • Thanks a lot, your last comment (especially the last sentence) helped A LOT! – Ovi Jun 05 '16 at 15:50