I don't really understand this problem, but I'm going to spill out what I've taken notes on. I know that in order to solve this we would need to use the contrapositive in each direction.

I'm going to use 'a' in place of phi

So we want to show that a is PL valid iff a is LP valid.

I think the next thing to do is assume a is not LP valid - I would also use Kleene's valuation function: there is a trivalent I, KVI(a)=0

In order to show a is not LP valid, I think we would need to prove that vi(a)=0 and KVi(a)=0.

To be honest, I don't know exactly what direction to take this. From class, we were told the crucial aspect was to prove Vi(a)=0 and KVi(a)=0

  • 2
    What is PL ? Prop logic ? and LP ? – Mauro ALLEGRANZA Feb 19 at 7:00
  • It seems that you are working with three-valued logics. there are many... – Mauro ALLEGRANZA Feb 19 at 9:01
  • I suggest you re-ask this on Mathematics (with the same tags). They will be more likely able to help you with a symbolic derivation/proof. – Mitch Feb 20 at 17:32
  • If the answer below is enough for you, please accept it and we can "close" the post. – Mauro ALLEGRANZA Feb 21 at 10:24

Since classical valuations (functions from atomic formulas to classical truth values) are LP-valuations it trivially follows that LP-validity implies classical validity.

For the converse assume that formula A is not LP-valid and so V(A) is not a designated value (it is 0 but neither 1 nor 1/2 in some representations of truth values), for some LP-valuation V. Now consider the function V' obtained from V by replacing all and only its 'middle values' 1/2 by 1 (truth only). More precisely, V'(p) = V(p), if V(p) is in {0,1} and V'(p) = 1 otherwise. V' is a classical valuation. A simple induction on the complexity of formulas shows that V and V' satisfy exactly the same formulas. So V'(A) = 0 and so A is not classically valid.

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