Lecture Notes, 1 Oct 2010
Last Update: 1 October 2010
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§1.5: Rules of Inference (cont'd)
- Rule of Inference vs. Tautologies:
- Recall that MP is the rule of inference:
& (A → B)
you may infer B
- It looks like the proposition:
(*) (A ∧ (A → B)) → B)
But it isn't!
- MP is a sequence of 3 propositions;
it is a valid argument form
- (*) is a single proposition;
it is a tautology
- But they are related:
MP is valid iff (*) is a tautology
- In general, a rule of inference
is valid iff ((A ∧ B) →
C) is a tautology
- This is called the "Deduction Theorem", and requires proof
(but we won't prove it)
- More Rules of Inference:
- Disjunctive Syllogism:
- Modul Tollens (method of denying (the
- There is a chart of some rules of inference on p. 66 (Table 1).
- In general, for each connective, there are "introduction" and "elimination" rules,
- some of which require nested subproofs.
- An "introduction" rule for a connective "*" tells you how to "introduce" as a line of a proof
a new proposition whose principal connective is "*".
- An "elimination" rule for a connective "*" tells you
how to "eliminate" a connective to create a new proposition for a line
of a proof.
- Rules of inference are "atomic" valid argument forms.
- Longer arguments are valid if their "sub-arguments" are
a sub-argument is valid if its sub-arguments are valid;
and so on, stopping at "atomic" arguments that are rules of inference.
- This is another example of what I have been calling
- The Resolution Rule of Inference:
- Resolution is a single rule of inference that can do
the work of all of the others.
- ∴, it's useful for computational implementations of
- The programming language Prolog is based on FOL and
- Consider the following chart of rules of inference.
- The familiar rules are in the first column.
- The second column rewrites the rule using only ¬ & ∨
(which we know are functionally complete!)
||A → B
||¬A ∨ B
||A → B
||¬A ∨ B
||A → B
B → C
A → C
||¬A ∨ B|
¬B ∨ C
¬A ∨ C
||A ∨ B
||(already uses only ¬,∨)
- Generalizing this pattern, we have Resolution:
A ∨ B1 ∨ … ∨ Bn
¬A ∨ C1 ∨ … ∨ Cm
(B1 ∨ … ∨ Bn)
(C1 ∨ … ∨ Cm)
or, more simply:
A ∨ B
¬A ∨ C
B ∨ C
- To use this as the only rule,
we need to represent all propositions in "clause form"
(or "conjunctive normal form")
i.e., using only ¬, ∧, & ∨
- Take CSE 463 to learn more.
- More examples of proofs:
- Prove that this argument is valid:
P1: Lynn works part time or full time.
P2: If Lynn doesn't play on the team,
then she doesn't work part time.
P3: If Lynn plays on the team,
then she's busy.
P4: Lynn doesn't work full time.
C : ∴ Lynn is busy.
- Syntax & Semantics of Representation:
Let pt = Lynn works part time.
Let ft = Lynn works full time.
Let plays = Lynn plays on the team.
Let busy = Lynn is busy.
- Translation of argument:
P1: pt &or ft
P2: ¬plays → ¬pt
P3: plays → busy
C : ∴ busy
- We want to prove that this is a valid argument
- without using truth tables
- How? Next time!
Text copyright © 2010 by William J. Rapaport
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