How to infer ¬Q when there seems to be no way to



Rule #1: No man shall hit another man.

Rule #2: If someone breaks Rule #1, then Rule #1 does not apply to such a one.

My specific question is: How can someone infer that Rule #1 does apply to him?

It’s not enough to say, “I am not breaking Rule #1—hence, it applies to me.” Such reasoning would be denying the antecedent, which is a formal fallacy. How can someone infer “¬Q” in this case?

London Jennings

Posted 2016-07-21T04:14:44.347

Reputation: 81

What do you imagine not Q to be? You never supply your formalization other than that single bit. – virmaior – 2016-07-21T04:31:31.667

@virmaior Well the P→Q statement is “If someone breaks Rule #1, then Rule #1 does not apply to such a one,” so ¬Q would be “Rule #1 does apply to [X].” – London Jennings – 2016-07-21T06:09:28.437

Try symbolizing it completely and adding that to your question by editing. (Note also that in sentential logic there's no way to include a variable in the symbolization in that way). – virmaior – 2016-07-21T07:43:35.553

1From P→Q alone, you cannot derive ¬Q (try with a valuation v such that v(Q)=true). – Mauro ALLEGRANZA – 2016-07-21T08:47:44.213



The way you have chosen to express the rules implies you are assuming a non-monotonic form of reasoning. Rule #1 as stated has no exceptions, while rule #2 expresses an exception to rule #1. In a monotonic system of logic (which includes classical logic) this would lead to a contradiction: if Bob hits Charlie, rule #1 says Charlie may not hit Bob back, but rule #2 says he may. In non-monotonic systems, rules may allow the inference of propositions that hold by default but may be defeated or overriden by the addition of other propositions. In such cases you would need some meta-rules that tell you how to apply the rules. For example, the rules might have some explicit priority value that tells you when one overrides another, or there might be a general consideration that more specific rules override general ones. In your example, rule #1 might then be assumed to hold by default but be defeasible where rule #2 applies, because rule #2 is more specific. You don't need to infer that the rule applies, you only need to check that no defeating conditions are present.

If you wished to avoid using non-monotonic reasoning, an alternative approach would be to attempt to express the obligation in a single rule, e.g. "no man shall hit another man who has himself never hit others". You can then infer that if Charlie is a man who has never hit others, then Charlie should not be hit.

The kind of reasoning we are using here is called deontic logic - the logic of obligation. Obligation can be treated as a propositional modality and attempts have been made to define formal logics for it, though it has proved highly problematic. The Stanford Encyclopedia has an article on deontic logic.


Posted 2016-07-21T04:14:44.347

Reputation: 11 885

So if I wrote: “no man shall hit another man ↔ if that man has never hit others,” I would be in the clear? – London Jennings – 2016-07-21T08:47:22.153

The propositional calculus isn't adequate here because you can't capture the meaning of "no man" or "that man" without quantification. I would rather express it using a dyadic obligation operator O(Q | P) meaning Q is obligatory in circumstances P. It might be expressed as (∀x)(∀y)O(¬Hits(x,y) | ¬(x=y) ˄ ¬(∃z)(Hits(y,z) ˄ ¬(y=z))) – Bumble – 2016-07-21T17:40:01.520

But you could just rewrite that as (∀x)(∀y)(¬(x=y) ˄ ¬(∃z)(Hits(y,z) ˄ ¬(y=z)) → O(¬Hits(x,y))), in which case you still run into the problem of denying the antecedent. – London Jennings – 2016-07-21T20:08:18.993

In general, a dyadic modal operator is not equivalent to the use of material implication: in fact one of the main reasons to use such operators is to avoid the problems that arise with material implication. But even if you use the formula you suggest, I don't see how it is an instance of denying the antecedent. If you instantiate the variable y with the constant c for Charlie, then provided the antecedent conditions hold, O(¬Hits(x,c)) follows by Modus Ponens. – Bumble – 2016-07-22T03:45:26.047