class 4 Flashcards
What we obtain by formalization
– Simplify natural language
– Eliminate ambiguities
– Capture the logical form
declarative sentences
Premises and conclusions in arguments, linguistic expressions that can be (said) true or false.
basic sentential operators (to form a complex sentence)
- Negation: not…
- Conjunction: … and …
- Disjunction: … or …
- Conditional: if … then …
- Bi-conditional: … if and only if …
and
not
–|
or
v
if
–>
if and only if, iff
<–>
God exists if and only if the cat is not on the sofa
q <–> –| p
God does not exist or the cat is not on the sofa
–| q v –| p
If the Prime Minister is on trial, he resigns, or the government loses
any credibility. The Prime Minister is on trial, and he does not resign,
so the government loses any credibility
p –> (q v r), p ^ –| q |- r
If Paul or Quinn go to the party, then Robert goes too
(p v q) –> r
It is not the case that Robert and Paul go to
the party
–| (r ^ p)
If Robert goes to the party, then if Paul goes then Quinn does not
r –> (p –> –| q)
–|p is true iff
p is false
p ^ q’ is true iff
both p and q are true
p v q is true iff
p or q is true, at least one
p | q is true iff
exclusive disjunction, at least p or q is true, but they cannot be both true
p –> q is true iff
it is not the case that the first
(antecedent) is true and the second (consequent) is false
–|(p ^ –|q)
p <–> q
p and q have the same truth value : both false or both true
(p –> q) ^ (q –> p)
If Robert goes to the party, then if Paul goes, then Quinn does not
r –> (p –> –|q)
Paul goes to the party, but Quinn does not
p ^ –|q
Neither Robert nor Quinn go to the party
–|r ^ –| q
Quinn goes to the party if Paul does not go
–|p –> q
Robert goes to the party only if Paul does not
r –> –| p
Paul goes to the party even if Quinn goes
p ^ q
Although Robert goes to the party, Paul does not go
r ^ –| p
Robert goes to the party provided that Paul goes
r –> p
Quinn goes to the party unless Robert goes
q v r which means that –|q –> r
Given that ‘p’ is true, ‘q’ is false, and ‘r’ is true, check the truth value
p ^ q
False, because ‘q’ is false (in case of ‘^’ both must be true)
Given that ‘p’ is true, ‘q’ is false, and ‘r’ is true, check the truth value
p –> q
False, because the antecedent is true but the consequent false (in case
of ‘→’ it is the only false case)
Given that ‘p’ is true, ‘q’ is false, and ‘r’ is true, check the truth value
–|p v –| q
True, because ‘q’ is false, so one of the two disjuncts, ‘–|q’, is true (in
case of ‘v’ that one is true is enough)
Given that ‘p’ is true, ‘q’ is false, and ‘r’ is true, check the truth value
q –> (p ^ –| r)
True because both, antecedent and consequent, are false
Given that ‘p’ is true, ‘q’ is false, and ‘r’ is true, check the truth value
(–|q ^ r) <–> p
True, because the two sides of the bi-conditional have the same T value
Given that ‘p’ is true, ‘q’ is false, and ‘r’ is true, check the truth value
(–| v r) –> –|q
True because both, antecedent and consequent, are true
Given that ‘p’ is true, ‘q’ is false, and ‘r’ is true, check the truth value
q –> (–|p<–> (–|r –> –|q))
true because it is possible for the antecedent to be negative and the consequent to be positive
