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16 Cards in this Set
- Front
- Back
In FOL
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Every name must name an object
No name can name more than one object An object can have more than one name or no name at all |
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Every predicate symbol comes with a single, fixed 'arity' a number that tells you how many names it needs to form an atomic sentence
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Every predicate is interpreted by a determinate property or relation of the same arity as the predicate
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Atomic sentences are formed by putting a predicate of arity "n" in front of "n" names
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Atomic sentences are built from the identity predicate = using infix notation: the arguments are placed on either side of the predicate
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The order of the names is crucial to forming atomic sentences
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Complex terms are typically formed by putting a function symbol of arity "n" in front of "n" terms
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Complex terms are used just like names in forming atomic sentences
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A proof of a statement S from premises P, Q is a step-by-step demonstration which shows that S must be true in any circumstances in which the premises P, Q are all true
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~P is true if and only if P is not true
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P^Q is true if and only if P is true and Q is true
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P v Q is true P is true or Q is true or both are true.
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parenthesis must be used whenever ambiguity would result from their omission. Conjunctions and disjunctions must be wrapped in parenthesis when combined by some other connective
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DeMorgans Laws
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Double negation ~~P=P
DeMorgan ~(P ^ Q) = (~P V ~Q) DeMorgan ~(P V Q) = (~P ^ ~Q) |
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S=sentence built of atomic sentences connected by truth functional connectives
Truth depends on truth of atomic parts |
Tautology=Every row assigns true to S
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Satisfiable=@ least one non-spurious row assigns True to S
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Logical true=every non-spurious row assigns true to S
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Proof by contradiction
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~S=Assume S and prove contradiction
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Distribution of ^ over V
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P ^ (Q V R) = (P^Q) V (P^R)
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Distribution of V over ^
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P V (Q ^ R) = (PVQ) ^ (PVR)
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Disjunctive Normal Form
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Disjunction of one or more conjunctions
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Conjunctive Normal Form
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Cconjunction of one or more disjunctions
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Any sentence can be in CNF or DNF or both
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P -> Q false only when P is true but Q is false
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