Does predicate logic have truth tables?

Question

As I recall in propositional logic, it was possible to draw truth tables for the arguments such as for:

(P ∨ R)   [I live in Paris or I live in Rome]
Therefore, (~P ⊃ R)  [If I don't live in Paris then I live in Rome]

You have a truth table given as:

   +---+----+---------+------------+
   | P | R  | (P ∨ R) |  (~P ⊃ R) |
   +---+----+---------+------------+
   | 1 | 1  |    1    |     1      |
   | 1 | 0  |    1    |     1      |
   | 0 | 1  |    1    |     1      |
   | 0 | 0  |    0    |     0      |
   +---+----+---------+------------+

But when you have argument in predicate logic such as:

~(∃x)Fx
Therefore, (x)~(Fx • Gx)

Can the similar form of truth table be derived to test for validity rather than a proof solving approach?

And, one of my other quick question is: Is it possible to convert the above argument (with P and R) given in Propositional logic into Predicate logic or it can only be written in propositional logic?

Answer 1

NO, because validity for predicate logic means true in all interpretations, and thus we have to take into account also interpretations with infinite domains, like the set N of natural numbers.

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Every tautology of propositional logic, like P ∨ ¬P, can produce an unlimited supply of valid predicate logic formulae through uniform substitution, i.e. by replacing every occurrence of a propositional letter by an atom of predicate logic language.

For example, from P ∨ ¬P we can produce the valid formulae :

∀xP(x) ∨ ¬∀xP(x)

∃xP(x) ∨ ¬∃xP(x)

and so on.

With your example, from P ∨ R ⊨ ¬P ⊃ R we can derive e.g. :

∀xP(x) ∨ ∃xQ(x) ⊨ ¬∀xP(x) ⊃ ∃xQ(x).

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But not all valid formulae of predicate logic are "substitution instances" of tautology; the formula ∀x(x=x) is valid but we can get it by uniform substitution only from the propositional logic formula P, that is not a tautology.

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Note

As per Owen's answer, we have to note that Monadic predicate calculus is a fragment of first-order logic that is decidable.

Answer 2

Does predicate logic have truth tables?

Yes! If we assume there are 4 function values (1,2,3,0) of Fx then monadic truth functions have truth tables to resolve expressions such as ~∃xFx -> ~∃x(Fx & Gx).

~1=0, ~2=3, ~3=2, ~0=1.

∃1=T, ∃2=T, ∃3=T, ∃0=F.

∀xFx =def ~∃x~Fx.

∀1=T, ∀2=F, ∀3=F, ∀0=F.

∀xFx -> ∃xFx, is tautologous.

Proof:

(∀1 -> ∃1) & (∀2 -> ∃2) & (∀3 -> ∃3) & (∀0 -> ∃0).

ie. (T -> T) & (F -> T) & (F -> T) & (F -> F).

(T) & (T) & (T) & (T).

That is ∀xFx -> ∃xFx, is true for all function values of Fx.

In the same way we can show that ~∃xFx -> ~∃x(Fx & Gx) is a tautology.