程序代写代做代考 Slide 1
Slide 1
Answers to Some Predicate Logic
Formalisation Exercises from the
Predicate Logic Notes
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Predicates to be used:
lecTheatre/1
office/1
contains/2
lecturer/1
has/2
same/2
phd/1
supervises/2
happy/1
completePhd/1
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1.
311 is a lecture theatre and 447 is an office.
311 is a lecture theatre and 447 is an office
lecTheatre(311) office(447)
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2. Every lecture theatre contains a projector.
Every lecture theatre contains a projector.
X ( → )
Every
Any Map to
All
Everyone
….
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Every lecture theatre contains a projector.
X (lecTheatre(X) → )
X (lecTheatre(X) → contains(X, projector))
Alternatively:
X (lecTheatre(X) →
Y (projector(Y) contains(X, Y)))
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A universally quantified wff is usually like this:
X ( → )
Or like this:
X ( )
That is their principle connective is
→ or .
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3. Every office contains a telephone and either a desktop or a laptop
computer.
Every office contains a telephone and either a desktop or a laptop computer.
X (office(X) → …….)
… contains a telephone and either a desktop or a laptop computer.
X (office(X) →( …. …))
X (office(X) →(contains(X, telephone) (…)))
… either a desktop or a laptop computer.
X (office(X) →(contains(X, telephone) ( … …)))
X (office(X) →(contains(X, telephone)
(contains(X, desktop) contains(X, laptop))))
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4. Every lecturer has at least one office.
Every lecturer has at least one office.
X (lecturer(X) → … )
… has at least one office.
X (lecturer(X) → Y (office(Y) has(X, Y)))
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At least one
Some map to
One
….
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An existentially quantified wff is usually like
this:
X ( )
Or like this:
X ( )
That is their principle connective is
or .
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5. No lecturer has more than one office.
No lecturer has more than one office.
Not any lecturer has more than one office.
(There does not exist a lecturer who has more
than one office.)
¬L (lecturer(L) L has more than one office)
¬L (lecturer(L) L has at least 2 offices that are
not the same)
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¬L (lecturer(L) there are at least 2 offices that
L has and are not the same)
¬L (lecturer(L)
O1 O2 (office(O1) office(O2)
has(L, O1) has(L, O2)
¬same(O1,O2)))
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There are other ways of doing 5. For example:
L (lecturer(L) → ¬ (O1 O2 (office(O1)
office(O2) has(L, O1) has(L, O2)
¬same(O1,O2)))
L O1 O2 (lecturer(L) office(O1)
office(O2) has(L, O1) has(L, O2) →
same(O1,O2))
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6. No lecturers share offices with anyone.
Try it …..
7. Some lecturers supervise PhD students and
some do not.
(L …..) (L …..) or
(L1 …..) (L2 …..)
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Some lecturers supervise PhD students and
some do not.
(L (lecturer(L) there is at least one PhD
student that L supervises))
(L (lecturer(L) there is not at least one PhD
student that L supervises))
(L (lecturer(L) S (phd(S) supervises(L,S))))
(L (lecturer(L) ¬S (phd(S) supervises(L,S))))
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(L (lecturer(L) S (phd(S) supervises(L,S))))
(L (lecturer(L) ¬S (phd(S) supervises(L,S))))
Drop some unneccesary brackets:
L (lecturer(L) S (phd(S) supervises(L,S)))
L (lecturer(L) ¬S (phd(S) supervises(L,S)))
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8. Each PhD student has an office, but all PhD
students share their office with at least one
other PhD student.
Try it ……
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9. A lecturer is happy if the PhD students
he/she supervises successfully complete their
PhD.
L (lecturer(L) his/her PhD students
successfully complete → happy(L))
L (lecturer(L) all his/her PhD students
successfully complete → happy(L))
L (lecturer(L) S if S is L’s PhD student then S
successfully completes→ happy(L))
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L (lecturer(L)
S if S is L’s PhD student then S successfully
completes→ happy(L))
L (lecturer(L)
S (phd(S) supervises(L,S) → completePhD(S))
→ happy(L))
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10.Not all PhD students complete their PhD.
Try it ….
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