Suppose $x \in R^+$ and we want to prove the implication $x < 0 \Rightarrow P(x)$, where $P(x)$ is some statement of $x$.

How should one tackle this situtation ?

Normally one should prove the implication in the case that the antecedent is false and in the case that the antecedent is true, where the false case follows (is true) by definition.

Should one prove the implication in the case $x < 0$ (antecedent is true) even thus $x \in R^+$, or is the case optional (we can just say the truth case will never happen ?).

More generally suppose we don't know whether $P(x)$ can be true for any $x$, and we want to prove $P(x) \Rightarrow Q(x)$.

Is it then O.K to assume $P(x)$ is true if it is false for all values of $x$ ? Should one prove the implication for $P(x)$ assumed to be true in order to prove the implication ?

  • $\begingroup$ I think I address your concerns here. $\endgroup$ – Git Gud Jun 2 '14 at 18:11
  • $\begingroup$ Would you comment on the comment I posted on @amWhy answer ? :) $\endgroup$ – Shuzheng Jun 2 '14 at 18:53

Since $x\in \mathbb R^+$, $$x \lt 0 \rightarrow P(x)$$ is always true, because an implication is always true when the antecedent if false.

Remember that the only situation in which an implication $a \rightarrow b$ is false is when $a$ (antecedent) is true AND $b$ (consequent) is false. If needed, refer to the truth-table for $\rightarrow$:

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Can you see that the only way an implication $a \rightarrow b$ can be false is if $a$ is true, and $b$ is false?

Since in our case, the antecedent $x \lt 0$ is false, it doesn't matter what $P(x)$ is, the implication as a whole doesn't meet the conditions of falsehood, so is thereby true.

In general:

$P(x)\rightarrow Q(x)$ (the implication as a whole) is certainly true if $P(x)$ is false for all values of $x$ in the domain, and this holds regardless of the truth value of $Q(x)$

  • $\begingroup$ Ahh thanks. So I should not break it up into two cases where I assume antecedent is false/true, and then prove the implication for each case, in the case that the antecedent is always false (in order to prove an implication)? What if I want to prove the implication in the case I don't know whether any value of $x$ makes the antecedent true ? $\endgroup$ – Shuzheng Jun 2 '14 at 18:09
  • $\begingroup$ If you know the consequent is true, the the implication is true. Otherwise, you cannot say. (See the truth table: whenever $b$ is true, the whole implication is true.) On another vein, if the antecedent of an implication is true, the implication is true if and only if the consequent is true. $\endgroup$ – Namaste Jun 2 '14 at 18:17
  • $\begingroup$ Suppose we want to prove Fermats Two Square theorem and wanted to prove the implication $$p = a^2 + b^2 \Rightarrow p \equiv 1 \pmod 4$$ in the case we haven't found any $a,b$ such that the antecedent could be true. Is it then illegal to assume the antecedent is true ? Suppose there aren't any $a,b$ satisfying the antecedent (I know there are, but suppose) and $p$ satisfies the consequence. Then we prove an implication with the assumption that the antecedent is true even thus it isn't ? Isn't this very illegal ? Should one always come up with at least one pair $a,b$ before proving ? $\endgroup$ – Shuzheng Jun 2 '14 at 18:39
  • $\begingroup$ In this case where we don't know whether there exist $a,b$, in order to prove the implication we need to assume it is true, that is there exist $a,b$ satisfying the antecedent ? If we could prove that none exist, the problem reduce to the original post-problem and the antecedent is false for every $a,b$. $\endgroup$ – Shuzheng Jun 2 '14 at 18:45
  • $\begingroup$ It is legitimate to try to deduce the conclusion by assuming the premise (antecedent). If we assume the premise is true, and can derive the conclusion through valid rules of inference, then we say that the conclusion follows from the premise, and we can conclude, then, that IF the premise/antecedent is true, then the conclusion must be true. Here we are talking more about an argument and its validity. It is true that valid argument forms hold regardless of the truth values of the propositions. $\endgroup$ – Namaste Jun 2 '14 at 19:00

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