INFERENCE RULES

(Sentential Rules)

We have 8 inference rules in our system.  They are:

 Rules of Inference Modus Ponens (MP) p → q p Therefore: q Modus Tollens (MT) p → q ~ q Therefore: ~p Hypothetical Syllogism (HS) p → q q → r Therefore: p → r Simplification (SIMP) p ● q Therefore: p Addition (ADD) p Therefore: p ▼ q Conjunction (CONJ) p q Therefore: p ● q Constructive Dilemma (CD) (p → q) ● (r → s) (p ▼ r) Therefore: (q ▼ s) Disjunctive Syllogism (DS) p ▼ q ~ p Therefore: q

Using inference rules works in much the same way as using equivalence rules.  We need to map the WFF's with which we are working onto premises in the argument form we wish to apply.  We must get an exact mapping, one WFF being a substitution instance of each premise in the argument form.  Once we have achieved this mapping, we generate a substitution instance of the conclusion of the argument form replacing the sentential variables in the conclusion with the WFF's that correspond to those variables in the premises.

Suppose that we wish to apply the inference rule Modus Ponens to the following set of WFF's:

(P Q) and P

The argument form for Modus Ponens is:

(p q)

p

q

so the mapping here is straightforward.

 Argument Form Argument (instance) (p → q) (P → Q) p P ∴q ∴Q

To take a less trivial example, we can apply Modus Ponens to the following set of WFF's:

[~(P Q) (~R  ~S)]  and ~(P Q)

The trick here is to see that the first WFF is a conditional whose antecedent is the negation of a conjunction and whose consequent is a conditional.  Here's how the mapping works:

 Argument Form Argument (instance) (p → q) [~(P ● Q) → (~R  → ~S)] p ~(P ● Q) ∴q ∴(~R  → ~S)

Now, if we also had the line ~R, we could apply Modus Ponens once again to that line plus the conclusion of the previous inference.

 Argument Form Argument (instance) (p → q) (~R  → ~S) p ~R ∴q ∴~S

Let's try one more example.  Let's apply Disjunctive Syllogism to the following WFF's:

~(P Q)  ▼ (R  ~S) and ~~(P Q)

Here's how the mapping works:

 Argument Form Argument (instance) p ▼ q ~(P ● Q)  ▼ (R  → ~S) ~p ~~(P ● Q) ∴q ∴(R  → ~S)

The key here is to see that ~(P Q) is a substitution instance of p in the argument form and that ~~(P Q) is an instance of ~p.   Pay close attention to the roles of negations is argument forms and in WFF's.  Often a negation in a WFF does not reflect a logical operation in the relevant argument form, as in this case.

WARNING: It is crucial that you remember that you cannot apply an inference rule to a part of a line.  Many students will try to simplify the antecedent of [(P Q)  R] in order to get (P R).  NO, NO, 1,000 times, NO!  To see why this inference does not work, imagine the following case--you need 2 more classes, logic and history, in order to graduate.  The registrar tells you that if you pass logic and history, then you will graduate.  It simply does not follow that if you pass logic, then you will graduate.