I am told that every tautological consequence is also a logical consequence, but what would a simple example be of a logical consequence that is not a tautological consequence?

Update To further explain what I have

  1. $Q$ is a tautological consequence of $P_1...P_n$ if and only if every row that assigns $True$ to each of $P_1...P_n$ also assigns $True$ to $Q$
  2. If $Q$ is a tautological consequence of $P_1...P_n$ then $Q$ is also a logical consequence of $P_1...P_n$
  3. Some logical consequences are not tautological consequences.

I understand on how to see if $Q$ is a tautological consequence from a truth-table but not sure on how to know if it is a logical consequence in the absence of a tautological one.

  • $\begingroup$ And what do you mean by logical consequence? $\endgroup$ – Gur Ismael Aug 19 '16 at 13:45

This situation only arises in first-order logic, not propositional logic. One formula is a tautological consequence of another if the two formulas arise by substituting first-order formulas into propositional formulas, so that the first propositional formula implies the second in propositional logic.

For example, the formula $$ A \land (A \to B) $$ implies the formula $$B \lor C$$ in propositional logic. Thus the first-order formula $$ (\forall x) S(x) \lor (\exists x) R(x) $$ is a tautological consequence of the formula $$ (\exists x)P(x) \land [(\exists x)P(x) \to (\forall x)S(x)]. $$ These are obtained by replacing the letters $A$, $B$, and $C$ with particular first-order formulas.

On the other hand, there are some logical implications in first-order logic that do not arise in this way. These implications tend to involve quantifier rules. For example, $(\exists x)S(x)$ is a logical consequence of $(\forall x)S(x)$ in first-order logic, but not a tautological consequence. Similarly $(\forall x)S(x)$ is a logical consequence of $(\forall y)S(y)$ but not a tautological consequence.


If I understand you correctly you speak about first order logic an by tautological consequence you mean e.g.

$$P(x)\wedge P(y)\models P(x),$$

in other word something that follows propositionally. And logical consequence would be e.g.

$$\forall x P(x)\models P(x),$$

where the inference involves quantifiers. Is this what you mean?

  • $\begingroup$ Updated question $\endgroup$ – darrrrUC Aug 19 '16 at 10:30

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