Logic with James B Nance

Conditional Proof and Reductio ad Absurdum

With the nine rules of inference and the ten rules of replacement taught in Lessons 13-17 of Intermediate Logic, we can construct a formal proof for any valid propositional argument. But for the benefit of the logic student, I introduce two additional rules in Lessons 18 and 19: the conditional proof, and the reductio ad absurdum. The conditional proof will often simplify a proof, especially one that has a conditional in the conclusion, making the proof shorter or easier to solve. The reductio ad absurdum method usually does not shorten a proof or make it that much easier to solve, but understanding the concept of reductio is beneficial for purposes outside of formal logic, such as understanding proofs in mathematics and apologetics.

Both conditional proof and reductio ad absurdum start with making assumptions. I want to clarify what happens with those assumptions.

To use conditional proof, you start by assuming the antecedent of a conditional. If you then deduce the consequent using that assumption and the other premises, you can conclude the entire conditional using conditional proof. The conditional proof reasoning can be symbolized like this:  p → q, ∴  p ⊃ q. In other words, if the assumed proposition p implies another proposition q, we can conclude if p then q.

One misconception that new logic students often make is thinking that the assumption actually “comes from” some previous step in the proof. They think that the assumption must appear somewhere else in order to make it. This is not the case. The assumed antecedent doesn’t come from anywhere; it is quite simply assumed. I tell my students we get the antecedent from our imagination; from Narnia, Middle Earth, Badon Hill. Using conditional proof, you are allowed to assume any antecedent you wish, as long as you use conditional proof properly from that point on.

A similar point can be made about reductio ad absurdum. To use the reductio method, you assume the negation of some proposition you want to conclude. If from that assumption and the other premises you deduce a self-contradiction, you can then conclude the original (un-negated) assumption. The reductio reasoning can be symbolized like this: ~p → (q • ~q), ∴ p. In other words, if assuming the negation of a proposition leads to a self-contradiction, we can conclude the proposition. But again, the assumption for the reductio does not need to appear elsewhere in the proof. We assume it “out of thin air.” Using reductio, you can assume anything you wish. If that assumption gives you a self-contradiction, you can conclude its opposite.

9 thoughts on “Conditional Proof and Reductio ad Absurdum

  1. Mr. Nance,
    I am very stumped on completing the following four problems and don’t know how to move forward on these:

    – Exercise 15a, #15
    – Exercise 17a, #8
    – Exercise 17a, #11
    – Exercise 17a, #12

    I would greatly appreciate any help you can give me. Thank you so much!

  2. Exercise 15a, #15: This one is challenging, because you have to use Absorption twice. Start by doing an Absorption on step 1 to get Y > (Y * Z). Then you should see the H.S. to do. Once you do the H.S., do Absorption again to set up a M.T.
    Exercise 17a, #8: You see that you have U > W for a premise, and W for the conclusion. This should show you that you need to convert ~(U > X) into U. Start with Impl. inside the parentheses, then using DeM.
    Exercise 17a, #11: You are trying to conclude ~H. To get it, you need to use M.T. (then Dem. etc.) on the first premise. To get that, you need to use M.P. on the second premise. To get that, you need to first turn K into J v K. This will take a couple of steps (starting with Addition).
    Exercise 17a, #12: You need to get ~P and ~L separately, then bring them together with Conj. on the last step. The ~L comes from steps 1 and 3 (with a couple additional steps), and the ~P from a M.T. on step 2 (along with a ~M from those same additional steps).

    I hope this helps. I am surprised you didn’t need assistance on Exercise 17a, #10. That’s the one that usually stumps my students.

  3. Thank you so very much for this clarification, Mr. Nance. We just covered this in our home school group and were very perplexed. Although this study of Logic proves challenging, it yields great satisfaction with each new discovery!

  4. Mr. Nance,

    Thank you so much! I will direct your answers to my son (this is his mom) tomorrow, in time for his CC Challenge B class. We’ve been raving about you, your blog, and your quick responses to the other students and parents. But then today couldn’t figure out how to actually post a question to you unless we “left a reply” to an already-posted question/answer. Essentially, how to direct a new question (new topic) to you.

    Thanks again!

  5. Hi Tonya, thank you for the question. Unfortunately, I have been unable to make that work. Supposedly WordPress can do that, but none of the experts in the office have been able to help me make it happen. For now, just keep checking back.

  6. Dear Mr. Nance,

    I just finished Intermediate Logic Lessons 18 (The Conditional Proof) and 19 (Reductio ad Absurdum) with my students.

    We certainly like using the conditional proof when possible, yet don’t really understand how and why it’s okay to assume the antecedent. Would you please shed some light on how or what is it that makes it okay to “assume” the antecedent and deduce the conclusion of a conditional in the conditional proof?

    I see where you symbolized the conditional proof reasoning as: p → q, ∴ p ⊃ q. In other words, if the assumed proposition p implies another proposition q, we can conclude if p then q.

    But, can you help us better understand WHY that is?

    Thank you very much,

    Julian Weber

    P.S. The Reductio ad Absurdum proof makes a bit more sense since it’s similar to what we did with determining equivalence, but the logic behind the conditional proof has us stumped! Looking forward to your feedback.

  7. Julian,

    Thank you for your question. In asking it, you have nearly given the answer yourself. You correctly summarized Conditional Proof as, “if the assumed proposition p implies another proposition q, we can conclude if p then q.” Recalling that if p then q can also be understood as “p implies q” (p. 31), then you are really saying, “if the assumed proposition p implies another proposition q, we can conclude that p implies q.” That an assumed proposition p can imply another proposition q is what if p then q means.

    The only other way I can defend Conditional Proof is to recognize that it works, meaning that when used properly it always results in a valid conclusion.

    Blessings,
    Mr. Nance

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