# User:JMRyan/Test1

### Modus Tollens

$\mathbf{T7.} \quad (\mathrm{P} \rightarrow \mathrm{Q}) \land \lnot \mathrm{Q} \rightarrow \lnot \mathrm{P}\,\!$

 1. $(\mathrm{P} \rightarrow \mathrm{Q}) \land \lnot \mathrm{Q}\,\!$ Assumption    $[(\mathrm{P} \rightarrow \mathrm{Q}) \land \lnot \mathrm{Q} \rightarrow \lnot \mathrm{P}]\,\!$
 2. $\mathrm{P}\,\!$ Assumption    $[\lnot \mathrm{P}]\,\!$ 3. $\mathrm{P} \rightarrow \mathrm{Q}\,\!$ 1 KE 4. $\mathrm{Q}\,\!$ 2, 3 CE 5. $\lnot \mathrm{Q}\,\!$ 1 KE
 6. $\lnot \mathrm{P}\,\!$ 2–5 NI
 7 $(\mathrm{P} \rightarrow \mathrm{Q}) \land \lnot \mathrm{Q} \rightarrow \lnot \mathrm{P}\,\!$ 1–6 CI

Now we use T7 to justify the following rule.

Modus Tollens (MT)
$(\varphi \rightarrow \psi)\,\!$
$\underline{\lnot \psi \quad \quad \ }\,\!$
$\lnot \varphi\,\!$

Modus Tollens is also sometimes known as 'Denying the Consequent'. Note that the following is not an instance of Modus Tollens, at least as defined above.

$\lnot \mathrm{P} \rightarrow \lnot \mathrm{Q}\,\!$
$\underline{\mathrm{Q} \quad \quad \quad \quad}\,\!$
$\mathrm{P}\,\!$

The premise lines of Modus Tollens are a conditional and the negation of its consequent. The premise lines of this inference are a conditional and the opposite of its consequent, but not the negation of its consequent. The desired inference here needs to be derived as below.

 1 $\lnot \mathrm{P} \rightarrow \lnot \mathrm{Q} \,\!$ Premise 2 $\mathrm{Q}\,\!$ Premise 3 $\lnot \lnot \mathrm{Q}\,\!$ 2 DNI 4 $\lnot \lnot \mathrm{P}\,\!$ 1, 3 CE 5 $\mathrm{P}\,\!$ 4 DNE

Of course, it is possible to prove as a theorem:

$(\lnot \mathrm{P} \rightarrow \lnot \mathrm{Q}) \land \mathrm{Q} \rightarrow \mathrm{P}\ .\,\!$

Then you can add a new inference rule—or, more likeley, a new form of Modus Tollens—on the basis of this theorem. However, we won't do that here.

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Stuff {{Edit2|User:JMRyan|stuff}} yeilds: Stuff edit stuff
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{{If||B|C}} yeilds: C
$\phi\,\!$      $\psi\,\!$

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