Validity in Polyadic QL
 

 The semantics for Polyadic QL introduces no new concepts. The only difference is that when we instantiate quantified propositions where we have multiple quantifiers, we must instantiate the quantifiers one at a time -- we peel them off one by one, whether we're doing this using our model universe method or our truth tree method.

We begin with the Model Universe Method. Let's look at the instantiation of
 

(x)(y)Fxy

in a domain of two individuals, {a, b}. We begin with the left-most quantifier, in this case the existential quantifier. Remember that we instantiate existentially quantified propositions as disjunctions of the individuals in the domain. So we have:

  (y)Fay v (y)Fby

as our first instantiation. Now we have a conjunction of universally quantified propositions. We instantiate the left side as follows:

  (Faa & Fab) v (y)Fby

and the right side:

  (Faa & Fab) v (Fba & Fbb)

Our original proposition is true in the domain {a,b} just in case this last truth-functional wff is true, and we can now check by using a truth-table or other method from PL.

We turn now to truth-trees, and we'll construct a truth tree for the argument:

  (x)(y) Lxy
  (x)(y) Lxy

which is a tranlsation of:

  Someone loves everyone.
  Everyone loves someone.

Let's look at the truth-tree bit by bit. First we set it up with the premises and the negation of the conclusion:

  (x)(y) Lxy`
  ~(x)(y) Lxy

Next, we'll move the negation of the second wff inside, and we'll do it in two stages:

  (x)(y) Lxy   
  ~(x)(y) Lxy     checked
  (x)~(y) Lxy     checked
  (x)(y)~Lxy  

Now we will instantiate the existentially quantified wffs, the two uncheck wffs:

  (x)(y) Lxy     checked
  ~(x)(y) Lxy     checked
  (x)~(y) Lxy     checked
  (x)(y)~Lxy     checked
  (y) Lay  
  (y)~Lby  

We instantiated the first existentially quantified wff with "a" and the second with "b". That leaves us with two univerally quantified wffs, and we instantiate them with every constant on the path. Doing that, we get:

  (x)(y) Lxy     checked
  ~(x)(y) Lxy     checked
  (x)~(y) Lxy     checked
  (x)(y)~Lxy     checked
  (y) Lay  
  (y)~Lby  
  Laa  
  Lab  
  ~Lba  
  ~Lbb  

Our tree is completed. It has just one path, but that path is open. Therefore our argument is invalid.

Here are some truth-tree problems in QL.

 

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