Deduction Help

Philosophy Forums Philosophy
1

I need some help with deduction and the Group I rules (modus ponens, modus tollens, chain argument, disjunctive argument… etc).

Take the following example:

  1. P–>Q
  2. ~P–>S
  3. ~Q

Deduct S.

The answer would be:

  1. P–>Q
  2. ~P–>S
  3. ~Q
  4. ~P 1,3 MT (modulus tollens) <---- Why would this even be considered a modulus tollen??
  5. S 2,4 MP (modulus ponens)

My question is…how did you arrive at 4 and 5? I know what modulus tollens and modulus ponens are…but how do I come to a deduction (how do I work towards the deduction)? Can’t you use another rule? Are you restricted to using modulus tollens and modulus ponens for the argument above?

2

I’m not sure what your difficulty is here. 1 and 3 just is MT, and 2 and 4 just is MP.

You can substitute some equivalent rules, but you’ll get a much longer proof.

3
  1. P–>Q
  2. ~P–>S
  3. ~Q
  4. ~P 1,3 MT (modulus tollens) <---- Why would this even be considered a modulus tollen??
  5. S 2,4 MP (modulus ponens)

What I don’t understand is, why am I doing “~P” in 4 to get “S” in 5.

If ~Q, then obviously P must be negated as well to ~P. BUT, why is that called modulus tollen?

4

2 says “if not P, then S.” So, having “not P” establishes “S”. Which is the conclusion you are looking for.

It’s “modus”, not “Modulus”. You can find the translation of these latin terms on Wikipedia.

5

I must start by pointing out that Modus Tollens(denying the consequent) is not necessarily true.

If you don’t understand how 4 and 5 are achieved then you do not understand the derivation rules being used.

They are rules of logic. If P then Q is assumed, and ~ Q is also assumed, then modus Tollens says you can derive ~P.

Modus Tollens is short for a more complex proof that utilizes negation introduction and conditional elimination(modus pollens). It is essentially short hand for:

1A P–>Q
2A~Q

3A |P…subAssumption
|__________________
3B |Q…1A,3A–>E(If p then Q and P then Q. We assume P to get Q)
3C |~Q…2A, R(repeating what has already been assumed
3D~P…3B,3C Negation introduction(since when I assume P I get the contradiction of Q and ~Q, I am entitled to say that P cannot be true, hence ~P)

Hence we derive ~P from the two original assumptions, and we get Modus Tollens.

You do not have to use modus Tollens, you can do the round about way, but it’s pointless to do so.

6

As Faust pointed out, you use modus Tollens because of the assumptions. You have to be able to intuit the way the proof is done based off of the assumptions. In more complex derivations this isn’t always that simple and may require many attempts. It gets easier when you memorize the rules and practice using them, but there should be a strong intuition within you telling you how to do such a problem.