Show that (p→q)→r and p→(q→r) are not equivalent
Im not sure what type of answer they want, i know im not supposed to use a truth table...
srry if this is a newbish question just started reading the textbook and im self teaching
Strictly speaking, one must use truth tables to prove that something is not equivalent in an introductory course. (There are also methods from proof theory, but I would think that they are not introductory-level.) Are there similar examples with solutions in the textbook? What is the context, i.e., what are the surrounding sections about?
To add: One way to show that two formulas are not equivalent is to convert them to full disjunctive normal forms and to show that they are different. However, the auxiliary statement that formulas are equivalent iff they have the same full disjunctive normal form is proved using truth tables.
Hello, whitetige1!
$\displaystyle \text{We know that }\,P \!\to Q\,\text{ and }\sim\!P \vee Q\,\text{ are equivalent.}$
$\displaystyle \text{Show that: }\:(p \to q) \to r \;\;\rlap{\;\:\:/}{\Longleftrightarrow} \;\; p \to (q \to r)$
. . $\displaystyle \begin{array}{cccc}\text{Left side} &&& \text{Right side} \\ \hline \\[-3mm] (p \to q ) \to r &&& p \to (q \to r) \\ (\sim\!p \vee q) \to r &&& \sim\!p \vee(q \to r) \\ \sim(\sim\!p \vee q) \vee r &&& \sim\!p \vee (\sim\!q \vee r) \\ \underbrace{(p \,\wedge \sim\!q) \vee r} &&& \underbrace{\sim\!p \,\vee \sim\!q \vee r} \end{array}$
. = . = . . $\displaystyle \uparrow \;\;\text{ not equivalent } \;\; \uparrow$
My apologies for being vauge, the chapter the question was in had just introduced logical equivalencies (De Morgans Laws...etc) so very introductory.
I found an answer, however i don't understand how/why they've answered it this way there were no examples answered this way and the previous questions were not answered this way either it just seems very out of context,
ANSWER:
These are not logically equivalent because when p, q, and
r are all false, (p→q)→r is false, but p→(q→r) is true.
maybe i should supplement the book im reading with something else.
Thanx for the quick responses
Hello, whitetige1!
I agree with Plato . . .
$\displaystyle \text{ANSWER:}$
$\displaystyle \text{These are not logically equivalent because when }p, q, r\text{ are all false,}$
. . $\displaystyle (p\to q)\to r\text{ is false, but }p \to(q\to r)\text{ is true. }\;(??)$
But how do we know this without using a truth table?
It just seems to be the way it was answered and the setting of the question that confused me, im not really fussed about it as i now understand it i just thought it would be worth asking to get some clarification. Thanks for all answers they are very much appreciated