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The logic of an implies :

$$\begin{cases}T \rightarrow T : T \\ T \rightarrow F : F \\ F \rightarrow T : T \\ F \rightarrow F : F \end{cases}$$

I am confused, when given an implies statement asking if its valid or not, technically, wouldn't F->F prove that the statement is valid?

Or is it only for T->T ?

Mauro ALLEGRANZA
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QOTDIANA
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    Your notation is confuse. –  Nov 08 '21 at 23:47
  • Note that $\rightarrow$ is not associative: $F \rightarrow (F \rightarrow F)$ but $(F \rightarrow F) \nrightarrow F$ – Henry Nov 09 '21 at 00:06
  • Is my "Latexifying" of your question what you mean ? Please note that, as underlined by Pat Bol, the fourth line should give "T" instead of "F". – Jean Marie Nov 09 '21 at 07:05

2 Answers2

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All combinations are true, except $$\color{red}{T\implies F}.$$

Note that in daily life, implications with a false antecedent ($F\implies F$ and $F\implies T$), which are true, are not considered.

Also note that in formal logic, the antecedent and consequent need not be related.

$$1>0\implies\pi\notin\mathbb Q$$ is a true statement.

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The following statements are true

  • $T \rightarrow T$
  • $F \rightarrow T$
  • $F \rightarrow F$

but $T \nrightarrow F$.

So (using brackets missing from your question) the following statements are also true

  • $(T \rightarrow T) \rightarrow T$
  • $(F \rightarrow T) \rightarrow T$
  • $(F \rightarrow F) \rightarrow T$
  • $(T \rightarrow F) \rightarrow T$
  • $(T \rightarrow F) \rightarrow F$

but $(T \rightarrow T) \nrightarrow F$ and $(F \rightarrow T) \nrightarrow F$ and $(F \rightarrow F) \nrightarrow F$

Henry
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