From Page 45 of Magnus/Button's book, ForAllx, the author(s) write

$A ⊨$

is equivalent to stating that the sentence $A$ is a contradiction.

The given definition of "$⊨$", a tautological entailment, is the following: The sentences $A_1, A_2,... ,A_n$ tautologically entail the sentence $C$ if there is no valuation of the atomic sentences which makes all of $A_1, A_2,... ,A_n$ true and $C$ false.

I am not sure how "$A ⊨$" can make sense, unless we admit that a statement of nothing is actually a sentence, but if we do, it seems that such a sentence must be both true and false, which we cannot admit in truth functional logic.

On the other hand "$⊨ C$" does make sense; it means that $C$ is a tautology. This is because (all the sentences on the right side are true) is a true statement, and if C could be false, the entailment does not hold. Therefore $C$ must be true for all valuations.

  • $\begingroup$ Well what about a contradiction isn't that both true and false? $\endgroup$ Apr 6, 2017 at 6:31
  • $\begingroup$ I think first one has to work out how this is a syntactically valid sentence itself. Does it implies entailment of a null statement? Does semantic entailment function as a binary and unary operation? $\endgroup$ Apr 6, 2017 at 6:39
  • $\begingroup$ Could be it is an arbitrary syntactic combination, meaningless in itself, chosen and given the defining trait of contradiction. $\endgroup$ Apr 6, 2017 at 6:42

1 Answer 1


Sequent calculus works with proof judgments of the form $$ A_1,A_2,\ldots,A_n \vdash B_1, B_2, \ldots B_k $$ which intuitively means, "if all of the $A_i$s are true, then at least one of the $B_i$s will be true too". Classically, this is the same as asserting that $$ \neg A_1 \lor \neg A_2 \lor \cdots \lor \neg A_n \lor B_1 \lor B_2 \lor \cdots \lor B_k $$ is a theorem (which makes the different treatment of commas on the left and right look a little less unmotivated). In particular, a proof of $$ A \vdash $$ in the sequent calculus proves that $A$ is never true, because there is nothing on the right there could be true when $A$ is.

In your case you have a $\vDash$ (which is a semantic notion) instead of $\vdash$, but it is still not completely crazy to extend the meaning of $\vDash$ to $k\ne 1$ formulas on the right in the same way as the extension works for $\vdash$. So

$$ A_1, A_2, \ldots A_n \vDash B_1, B_2, \ldots, B_k $$ should mean that for every valuation (interpretation, structure, model) that satisfies all of the $A_i$s true, there is at least one $B_i$ that it satisfies.

In particular $A_1,A_2,\ldots,A_n\vDash\varnothing$ will assert that there is no valuation that satisfies all of the $A_i$s.


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