1. Mathematical Structures for Computer Science Chapter 1
Formal Logic
Mathematical Structures for Computer Science
Chapter 1
Copyright © 2006 W.H. Freeman & Co. MSCS Slides Formal Logic

2. Four new inference rules
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
Basic approach to prove arguments:
Strip of quantifiers
Manipulate the unquantified wffs
Reinsert the quantifiers
Four new inference rules
Note to remember: P(x) could be (y) (z) Q(x,y,z)
Two rules to strip the qualifiers
Two rules to reinsert the qualifiers
Section 1.4
Predicate Logic

3. Inference Rules From Can Derive Name / Abbreviation
Restrictions on Use
(x)P(x)
P(t) where t is a variable or constant symbol
Universal Instantiation- ui
If t is a variable, it must not fall within the scope of a quantifier for t
(x)P(x)
P(t) where t is a variable or constant symbol not previously used in a proof sequence
Existential Instantiation- ei
Must be the first rule used that introduces t
P(x)
Universal Generalization- ug
P(x) has not been deduced from any hypotheses in which x is a free variable nor has P(x) been deduced by ei from any wff in which x is a free variable
P(x) or P(a)
Existential Generalization- eg
To go from P(a) to (x)P(x), x must not appear in P(a)
Section 1.4
Predicate Logic

4. Examples: Proofs using Predicate Logic
Prove the following argument:
All flowers are plants. Sunflower is a flower. Therefore, sunflower is a plant.
P(x) is “ x is a plant”
a is a constant symbol (Sunflower)
Q(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:
(x)[P(x)  Q(x)] hyp
P(a) hyp
P(a)  Q(a) 1, ui
Q(a) 2, 3, mp
Section 1.4
Predicate Logic

5. Examples: Proofs using Predicate Logic
Prove the argument (x)[P(x)  Q(x)] Λ [Q(y)]  [P(y)]
Proof sequence:
(x)[P(x)  Q(x)] hyp
[Q(y)] hyp
P(y)  Q(y) 1, ui
[P(y)] 2, 3, mt
Prove the argument (x)P(x)  (x)P(x)
(x)P(x) hyp
P(x) 1, ui
(x)P(x) 2, eg
Section 1.4
Predicate Logic

6. Examples: Proofs using Predicate Logic
Prove the argument (x)[P(x)  Q(x)] Λ (x)P(x)  (x)Q(x)
Proof sequence:
(x)[P(x)  Q(x)] hyp
(x)P(x) hyp
P(x)  Q(x) 1, ui
P(x) 2, ui : no restriction on ui about reusing a name
Q(x) 3, 4, mp
(x)Q(x) 5, ug
Note: step 6 is legitimate since x is not a free variable in any hypothesis nor was ei used before
Section 1.4
Predicate Logic

7. Examples: Proofs using Predicate Logic
Prove the argument
(x)[P(x) Λ Q(x)]  (x)P(x) Λ (x)Q(x)
Proof sequence:
(x)[P(x) Λ Q(x)] hyp
P(x) Λ Q(x) 1, ui
P(x) 2, sim
Q(x) 2, sim
(x)P(x) 3, ug
(x)Q(x) 4, ug
(x)P(x) Λ (x)Q(x) 5, 6, con
Section 1.4
Predicate Logic

8. Examples: Proofs using Predicate Logic
Prove the argument
(y)[P(x)  Q(x,y)]  [P(x)  (y)Q(x,y)]
Using the deduction method, we can derive
(y)[P(x)  Q(x,y)] Λ P(x)  (y)Q(x,y)
Proof sequence:
(y)[P(x)  Q(x,y)] hyp
P(x) hyp
P(x)  Q(x,y) 1, ui
Q(x,y) 2, 3, mp
(y)Q(x,y) 4, ug
Section 1.4
Predicate Logic

9. Temporary hypotheses
A temporary hypothesis can be inserted into a proof sequence. If T is inserted as a temporary hypothesis and eventually W is deduced from T and other hypotheses, then the wff T  W has been deduced from other hypotheses and can be reinserted into the proof sequence
Prove the argument
[P(x)  (y)Q(x,y)]  (y)[P(x)  Q(x,y)]
Proof sequence:
P(x)  (y)Q(x,y) hyp
P(x) temporary hypothesis
(y)Q(x,y) 1, 2, mp
Q(x,y) 3, ui
P(x)  Q(x,y) temp. hyp discharged
(y)[P(x)  Q(x,y)] 5, ug
Section 1.4
Predicate Logic

10. More Examples Prove the sequence [(x)A(x)]  (x)[A(x)]
To prove equivalence, implication in each direction should be proved
Proof sequence for [(x)A(x)]  (x)[A(x)]
[(x)A(x)] hyp
A(x) temp. hyp
(x)A(x) 2, eg
A(x)  (x)A(x) temp. hyp discharged
[A(x)] 1, 4, mt
(x)[A(x)] 5, ug
Proof sequence for (x)[A(x)]  [(x)A(x)]
(x)[A(x)] hyp
(x)A(x) temp. hyp
A(a) 2, ei
[A(a)] 1, ui
[(x)[A(x)]] 3, 4, inc
(x)A(x)  [(x)[A(x)]] temp. discharged
[((x)[A(x)])] 1, dn
[(x)A(x)] 6, 7, mt
Section 1.4
Predicate Logic

11. Proving Verbal arguments
Every crocodile is bigger than every alligator. Sam is a crocodile. But there is a snake, and Sam isn’t bigger than that snake. Therefore, something is not an alligator.
Use C(x): x is a crocodile; A(x): x is an alligator, B(x,y): x is bigger than y, s is a constant (Sam), S(x): x is a Snake
Hence prove argument
(x) (y)[C(x) Λ A(y)  B(x,y)] Λ C(s) Λ (x)(S(x) Λ [B(s,x)])  (x)[A(x)]
(x) (y)[C(x) Λ A(x)  B(x,y)] hyp
C(s) hyp
(x)(S(x) Λ [B(s,x)]) hyp
(y)[C(s) Λ A(y)  B(s,y)] 1, ui
S(a) Λ [B(s,a)]) 3, ei
C(s) Λ A(a)  B(s,a) 4, ui
[B(s,a)] , sim
[C(s) Λ A(a)] 6, 7, mt
[C(s)] V [A(a)] 9, De Morgan
[[C(s)]] , dn
[A(a)] , 10, ds
(x)[A(x)] 11, eg
Section 1.4
Predicate Logic

12. Class Exercise Prove the argument
(x)[(B(x) V C(x))  A(x)]  (x)[B(x)  A(x)]
Proof sequence:
(x)[(B(x) V C(x))  A(x)] hyp
(B(x) V C(x))  A(x) 1, ui
B(x) temp. hyp
B(x) V C(x) 3, add
A(x) 2, 4, mp
B(x)  A(x) temp. hyp discharged
(x)[B(x)  A(x)] 6, ug
Section 1.4
Predicate Logic

13. Class exercise
Every ambassador speaks only to diplomats, and some ambassadors speak to someone. Therefore, there is a diplomat.
Use A(x): x is an ambassador; S(x,y): x speaks to y; D(x): x is a diplomat
Prove the argument
(x) (y)[(A(x) Λ S(x,y))  D(y)] Λ (x)(y)(A(x) Λ S(x,y))  (x)D(x)
Proof Sequence:
(x) (y)[(A(x) Λ S(x,y))  D(y)] hyp
(x)(y)(A(x) Λ S(x,y)) hyp
(y)((A(a) Λ S(a,y))  D(y)) 1, ui
(y)(A(a) Λ S(a,y)) 2, ei
A(a) Λ S(a,b) 4, ei
(A(a) Λ S(a,b))  D(b) 3, ui
D(b) 5, 6, mp
(x)D(x) 7, eg
Proof Sequence:
(x) (y)((A(x) Λ S(x,y))  D(y)) hyp
(x)(y)(A(x) Λ S(x,y)) hyp
(y)((A(a) Λ S(a,y))  D(y)) 1, ui
(y)(A(a) Λ S(a,y)) 2, ei
A(a) Λ S(a,b) 4, ei
(A(a) Λ S(a,b))  D(b)) 3, ui
D(b) 3, 4, mp
(x)D(x) 7, eg
Section 1.4
Predicate Logic