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Suppose we have a logical system $s$. Now, the monotonicity property tells us that: $\Gamma \vdash A$ and $\Gamma \subseteq \Delta$ implies that $\Delta \vdash A$. I see this definition somewhat problematic: What if $\Gamma = \{A\}$ where $A = \text{"Socrates likes chocolate"}$. and $B = \text {"Socrates hates chocolate"}$. Now, it doesn't true that $\Gamma \cup \{B\} \vdash A$. Is it? I mean, $\Gamma \cup \{B\}$ is always a false set of statements.

What am I missing in this definition of monotonicity?

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  • $\begingroup$ Adding sentences does not invalidate the already existing proof, does it? You can just ignore them. $\endgroup$
    – Henno Brandsma
    Jun 5, 2016 at 14:09
  • $\begingroup$ Given some $A$: Semantically, $A$ should be true whenever $\Delta$ is true, but the later is never true, so isn't is a contradiction for completeness? because $\Delta\vdash A$ (A is provable from $\Delta$) but $\Delta \not\vDash A$ (A cannot be inferred from $\Delta$). $\endgroup$
    – LiorGolan
    Jun 5, 2016 at 14:21
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    $\begingroup$ No. Whenever $\Delta$ is true (which is never), then $A$ is true is a true statement. We cannot refute it: then we'd need a model where $\Delta$ holds but $A$ does not. If you cannot refute it, it's true. It's just like the implication $p \rightarrow q$ where $p$ is false, or $\forall_x p(x)$ where $x$ ranges over the empty set. Void truth. $\endgroup$
    – Henno Brandsma
    Jun 5, 2016 at 14:23
  • $\begingroup$ Oh I think I understand now. It's somewhat like everything is true from the empty set. $\endgroup$
    – LiorGolan
    Jun 5, 2016 at 14:29
  • $\begingroup$ Yes it is. But it's good to consider such things. $\endgroup$
    – Henno Brandsma
    Jun 5, 2016 at 14:30

1 Answer 1

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This property is correct. In fact, if $\Gamma$ is a set of sentences that leads to a contradiction, then $\Gamma\vdash A$ for every sentence $A$. In plain words, from a false statement we can deduce anything.

Proof: Let's say that $\Gamma\vdash B\wedge\neg B$. To prove $A$, let's suppose $\neg A$. Now we deduce (because we can) $B\wedge\neg B$ from the sentences in $\Gamma$. This is a contradiction, so $\neg A$ is false and hence $A$ is true.

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  • $\begingroup$ You meant $\Gamma = B\land \lnot B$? $\endgroup$
    – LiorGolan
    Jun 5, 2016 at 14:06
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    $\begingroup$ @LiorGolan No, he does mean $\Gamma$ proves a contradiction. $\endgroup$
    – Henno Brandsma
    Jun 5, 2016 at 14:11

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