User:JMRyan/Test2

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.