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i have problem solving the following exrecise in many valued logic given the many valued logic $ L_5 $ defined as follows

$ S = \{0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1\} $ the possible truth values
$ D = \{1\} $ the designated set

and the following truth function
$ \to (a,b) = \begin{cases} 1 & a \leq b \\ b & a>b \end{cases} $

a) is $ ((p\to q)\to p) \to p $ a tautology?
this is false using the followin assignment $ v[p] = \frac{1}{4} ,v[q] = 0 $
$ v[((p\to q)\to p) \to p] = \frac{1}{4} $
thus it isn't a tautology

the problem is in the following section:
b) prove that using the following proof system:
axioms:
I1. $ A \to (B \to A) $
I2. $ (A \to (B \to C)) \to ((A \to B) \to (A \to C)) $

inference rule
$ A, A \to B $ infer $B$
is not enough to prove any tautology in the classic propositional calculus (true and false possible truth values) use section a

i have no idea how to prove b

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1 Answer 1

up vote 3 down vote accepted

Hint: I1 and I2 are tautologies in $L_5$. Also modus ponens (the inference rule in you system) when applied to tautologies in $L_5$, produces a tautology in $L_5$. Hence everything you derive in your proof system is a tautology in $L_5$. So you need only to find a formula that is a tautology in the classical propositional logic, but not a tautology in $L_5$.

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