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For example if I have a logic that has a syntax and semantics but no inference rules does that make a logic complete but not sound since soundness states that $LHS \vdash RHS$ implies $LHS \models RHS$. But I'm guessing if we have some type of syntax and semantics we could construct some sort of proof or inference rules from the model and hence the logic it's complete.

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Suppose you have a logic $\mathit{Tiny}$ with two axioms:

$A1$: Socrates is a man (i.e. Socrates $\in$ Men)

$A2$: All men are mortal (i.e. $x \in \text{Men} \Rightarrow x \in \text{Mortals}$)

but no rules of inference.

The only theorems in $\mathit{Tiny}$ are $A1$ an $A2$ - without rules of inference you cannot derive any other theorems apart from the axioms which are given. $A1$ and $A2$ are also true, so $\mathit{Tiny}$ is sound.

However, $\mathit{Tiny}$ is not semantically complete because the statement "Socrates is mortal" is semantically true, but is not a theorem of $\mathit{Tiny}$.

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  • $\begingroup$ MathJax remark. Note the strange spacing in $Tiny$ .... Better to do it $\mathit{Tiny}$. That is, not $Tiny$ but $\mathit{Tiny}$. The automatic spacing $Tiny$ is what you want when you multiply four items $T,i,n,y$. $\endgroup$
    – GEdgar
    Oct 8, 2019 at 12:28
  • $\begingroup$ @GEdgar Thanks - fixed. $\endgroup$
    – gandalf61
    Oct 8, 2019 at 12:34
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    $\begingroup$ I have always wondered as a mere mathematician why we don't need a Rule which says that we may quote any axiom. Otherwise there are two sorts of theorem, axioms and those that need a proof - and this is messy. Perhaps @gandalf61 can enlighten me: his proof above depends on this silent convention. $\endgroup$ Oct 8, 2019 at 12:43
  • $\begingroup$ What if A1 or A2 is not true. Then would $\mathit{Tiny}$ not be sound? $\endgroup$
    – KetDog
    Oct 8, 2019 at 15:41
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    $\begingroup$ @SimonGarfe Indeed. <A1=Socrates is a man; A2=Socrates is not a man> is not even sound. $\endgroup$
    – gandalf61
    Oct 8, 2019 at 15:44

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