One of the valid forms of argument is Modus Tollens (ie If P, then Q. Not Q, therefore, not P). An example is "If Putnam is guilty, she is lying now. She is not lying now. Therefore Putnam is not guilty." (The Elements of Reasoning - R Munson & A Black 2012). Modus Tollens can be rearranged to: If not P then not Q, Q, therefore P. What is it that distinguishes Modus Tollens from Affirming the Consequent, which is invalid? Aren't both acknowledging the presence, or lack thereof of the consequent (Q) and then drawing a conclusion about P based on this. If so, then shouldn't Modus Tollens be classified as an invalid form of argument?

1 Answer 1


Don't be misled by the second way of presenting Modus Tollens (MT). The general form is:

MT:   {P → Q, ¬Q} ⊢ ¬P.

Your second form, call it MT′, is very different:

MT′:   {¬P → ¬Q, Q} ⊢ ;P.

From a natural deduction standpoint with a rule of conditional introduction and detachment (also called modus ponendo ponens), MT' will suffice for all negation and conditional tautologies of classical logic. But, MT all with a rule of conditional introduction and detachment can't produce all negation and conditional tautologies.

The fallacious inference of affirming the consequent (☆) has the following slightly different form:

(☆)   {P → Q, Q} ⊢ P.

This is not a substitution instance of MT, i.e., there is no assignment of values to P and Q in MT (or MT′) that will yield (☆).

Without going into the philosophical logical literature on MT, I'd like to convince you that MT is a fine rule of inference by unpacking the meaning of the material conditional (→), revealing another form of MT that you didn't mention in your question:

MT′′:   {¬P ∨ Q, ¬Q} ⊢ ¬P.

This form of MT is an instance of a general rule of inference called disjunctive syllogism (DS):

DS:   {P ∨ Q, ¬Q} ⊢ P.

MT′′ can be obtained from DS by substituting ¬P for P. Intuitively, consider the following assignments for P and Q:

  • (P) a is an even number,
  • (Q) a is an odd number,

for some arbitrary number a. DS says that, if you've proved that a is either an even number or an odd number, and you have further proved that a is not an odd number, then you have a proof for the fact that a is an even number.

If you find this reasoning acceptable (do you?), then you have to grant that MT is also an acceptable rule of inference.

To convince yourself that ☆ is not acceptable, let's build a countermodel:

  • (P) b is an even number,
  • (Q) b is a number,

for some object b (not necessarily a number). The compound formula (P → Q) says that "if b is an even number then b is a number". The inference says: if you have proven that if b is an even number then b is a number, and further you have proven that b is a number, then you have in fact proven that b is an even number.

This is clearly fallacious. Suppose b = 3. (P → Q) is trivially true on logical grounds. If we have proved Q then we know that 3 is a number, which is great to verify. But ☆ is allowing us to conclude that we've proved P, namely that: b (which is equal to 3) is an even number!

Therefore, we conclude that ☆ is a fallacious rule of inference, while MT is fine.

A lot of philosophical logical literature has been devoted to discussing the pros and cons of MT and similar basic inference rules and their different forms, and the principles we commit to when we accept them. I can't go into those because

  • (i) I'm not qualified, and
  • (ii) I find MT unproblematic (do you? if 'no' → countermodel?).
  • MT' is not a substitution instance of MT. Carrying of the substitutions yields {($\lnot$P $\to$ $\not$Q), $\lnot$$\lnot$Q} $\vdash$ $\lnot$$\lnot$P. Sep 19, 2016 at 2:11
  • 1
    @DougSpoonwood Actually, you had written \notQ instead of \lnot Q, so the negation wasn't being rendered. What you have there and MT' are the same thing, you just have the double negations to cancel out. I've invited you to the chat so that we can discuss this; comments are too slow. Sep 19, 2016 at 3:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .