Propositions, Equivalence and Rules of Inference
We remind ourself about propositional equivalence: First, one rule: "Contrapositive" a --> b <==> ~b --> ~a ~a v b <==> ~b --> ~a "implies rule" ~a v b <==> ~~b v ~a "implies rule" ~a v b <==> b v ~a "double negation" ~a v b <==> ~a v b "commutative rule" Second, can we add to the list of nice rules about "--->" Prove or disprove the "associative rule of implication". (p --> q) --> r <==> p --> (q --> r) ~(~p v q) v r ( we're just seeing if it works... so don't have to do all the steps ) (p ^ ~q) v r <==> ~p v (~q v r) this looks alllll different. so find some way of assigning t,f to all the variables. then go back and check in the very first line...
Going beyond Propositional Equivalence
So far, we've defined propositions, defined how to make compound expressions with connectives, talked about how to find logically equivalent propositions, and talked about quantifying statements so that we can make statements abot things that are true for all elements of a set.
ALL of these, don't really let you DO anything --- you are either writing down formally some compound proposition, or translating one way of representing a compound proposition with another compound proposition which is completely equivalent. The next bit starts to get more exciting. Suppose you have a list of facts. Perhaps you can determine that some new fact, which is NOT logically equivalent to the given list, must be true (if the whole list is true). This is called an argument.
A "argument" can be written as follows:
P Q R ----- S
Where P, Q, R (and perhaps more) are called "Hypotheses", and they may be more complicated that a single proposition. and S is the conclusion.
Definition of valid argument If (P^Q^R) --> S is a tautology, then this is a valid argument.
Proving that an argument is valid can be done using logical equivalences and "rules of inference" which are themselves simple arguments.
You can also prove an argument is valid using truth tables. How?
So lets introduce the rules of inference:
Rules: p --- p v q addition p^q --- p simplification p q --- p^q conjunction p p --> q ------- q modus ponens ~q p --> q ------- ~p modus tollens. (prove this one, use contrapositive, MP) p --> q q --> r ------- p --> r hypothetical syllogism (prove this one) p v q ~p ----- q disjunctive syllogism p v q ~p v r ----- q v r Resolution (Core of many automated reasoning systems) but we can prove it without resolution... p v q == ~p --> q using implies rule: ~p v r == r v ~p == ~r --> ~p using commutative and implies. ~r --> ~p ^ ~p --> q == ~r --> q == r v q using hypothetical syllogism...
Best of Luck
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