According to the precedence of logical connectives, operator $\rightarrow$ gets higher precedence than $\leftrightarrow$ operator. But what about associativity of $\rightarrow$ operator?
The implies operator ($\rightarrow$) does not have the associative property. That means that $(p \rightarrow q) \rightarrow r$ is not equivalent to $p \rightarrow (q \rightarrow r)$. Because of that, the question comes op how $p \rightarrow q \rightarrow r$ should be interpreted.
The proposition $p \rightarrow q \rightarrow r$ can be defined in multiple ways that make sense:
- $(p \rightarrow q) \rightarrow r$ (left associativity)
- $p \rightarrow (q \rightarrow r)$ (right associativity)
- $(p \rightarrow q) \land (q \rightarrow r)$
Which one of these definitions is used?
I could not locate any book/webpage that mentions about associativity of logical operators in discrete mathematics.
Please also cite the reference (book/reliable webpage) that you use to answer my question (as I'm planning to add this to wikipedia page about 'logical connectives').
Thanks.
PS: I got this question when I saw this problem:
Check if following compound proposition is tautology or not:
$$ \mathrm{p} \leftrightarrow (\mathrm{q} \wedge \mathrm{r}) \rightarrow \neg\mathrm{r} \rightarrow \neg\mathrm{p}$$
Best Answer
When you enter $p \Rightarrow q \Rightarrow r$ in Mathematica (with
p \[Implies] q \[Implies] r), it displays $p \Rightarrow (q \Rightarrow r)$.That makes it plausible that the $\rightarrow$ operator is generally accepted as right-associative.