Modus Tollendo Tollens/Sequent Form

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Proof Rule

The Modus Tollendo Tollens can be symbolised by the sequent:

$p \implies q, \neg q \vdash \neg p$


Proof 1

By the tableau method of natural deduction:

$p \implies q, \neg q \vdash \neg p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 2 $\neg q$ Premise (None)
3 1, 2 $\neg p$ Modus Tollendo Tollens (MTT) 1, 2

$\blacksquare$


Proof 2

By the tableau method of natural deduction:

$p \implies q, \neg q \vdash \neg p$
Line Pool Formula Rule Depends upon Notes
1 1 $p \implies q$ Premise (None)
2 2 $\neg q$ Premise (None)
3 3 $p$ Assumption (None) Assume $p$ ...
4 1, 3 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 3 ... and derive $q$ ...
5 1, 2, 3 $\bot$ Principle of Non-Contradiction: $\neg \mathcal E$ 4, 2 ... demonstrating a contradiction
6 1, 2 $\neg p$ Proof by Contradiction: $\neg \mathcal I$ 3 – 5 Assumption 3 has been discharged

$\blacksquare$


Proof 3

We apply the Method of Truth Tables to the proposition.

As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the left hand side is $T$, that under the one on the right hand side is also $T$:


$\begin{array}{|cccccc||cc|} \hline (p & \implies & q) & \land & \neg & q & \neg & p \\ \hline F & T & F & T & T & F & T & F \\ F & T & T & F & F & T & T & F \\ T & F & F & F & T & F & F & T \\ T & T & T & F & F & T & F & T \\ \hline \end{array}$

Hence the result.

$\blacksquare$


Note that the two formulas are not equivalent, as the relevant columns do not match exactly.


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