Lecture 1-4: Predicate Logic
Predicate Logic
• Similar to propositional logic for solving arguments, build
from quantifiers, predicates and logical connectives.
• A valid argument for predicate logic need not be a
tautology.
• The meaning and the structure of the quantifiers and
predicates determines the interpretation and the validity
of the arguments
Inference Rules
From
Can Derive
Name /
Abbreviation
(x)P(x)
P(c) where c is a
variable or constant
symbol
Universal
Instantiation- ui
(x)P(x)
P(c) for some
element c
Existential
Instantiation- ei
P(c) for an
arbitrary c
(x)P(x)
Universal
Generalizationug
P(c) for
some
element c
(x)P(x)
Existential
Generalizationeg
Proofs using Predicate Logic
•
Prove the following argument:
– All flowers are plants. Sunflower is a flower. Therefore,
sunflower is a plant.
•
•
•
•
•
Q(x) is “ x is a plant”
a is a constant symbol (Sunflower)
P(x) is “x is a flower”
The argument is (x)[P(x)  Q(x)] Λ P(a)  Q(a)
The proof sequence is as follows:
1.
2.
3.
4.
(x)[P(x)  Q(x)]
P(a)
P(a)  Q(a)
Q(a)
hyp
hyp
1, ui
2, 3, mp
More Examples
•
•
Prove the argument (x)[P(x)  Q(x)] Λ [Q(y)]  [P(y)]
Proof sequence:
1.
2.
3.
4.
•
•
(x)[P(x)  Q(x)]
[Q(y)]
P(y)  Q(y)
[P(y)]
hyp
hyp
1, ui
2, 3, mt
Prove the argument (x)P(x)  (x)P(x)
Proof sequence:
1.
2.
3.
(x)P(x)
P(c) for an arbitrary c
(x)P(x)
hyp
1, ui
2, eg
More Examples
•
Prove the argument
(x)[P(x) Λ Q(x)]  (x)P(x) Λ (x)Q(x)
•
Proof sequence:
1.
2.
3.
4.
5.
6.
7.
(x)[P(x) Λ Q(x)]
P(c) Λ Q(c) for an arbitrary c
P(c) for an arbitrary c
Q(c) for an arbitrary c
(x)P(x)
(x)Q(x)
(x)P(x) Λ (x)Q(x)
hyp
1, ui
2, sim
2, sim
3, ug
4, ug
5, 6, con
Dealing with Negation
•
[(x)A(x)]  (x)[A(x)]
•
(x)[A(x)]  [(x)A(x)]
Group Exercises
•
Formulate the argument in predicate
logic and then provide a proof
1. All M are P. All S are M. Therefore, all S are
P.
2. All M are P. Some S are M. Therefore, some
S are P.
3. No M are P. Some S are M. Therefore,
some S are not P.
– End of Lecture 1-4
– End of Chapter 1