Validity in Polyadic QL
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(∃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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