Introduction to Logic
Duration: 1h 15m
2005 / 06 – 4th Test
Nome:
nº:
1. Considering the predicates of the Language of the Tarski Worlds, translate into that language the
following numerical propositions, using only standard universal and existential quantification.
a) There are exactly 2 cubes.

x y ( Cube(x)  Cube(y)  x  y  z ((x  z  y  z)   Cube(z)))
b) Between any pair composed of a tetredron and a dodecaedron, there is 1 (and only 1) block.

xy((Tet(x)  Dodec(y)) z( Between(z,x,y) w(Between(w,x,y) z = w)))
c) No tetraedron is to the left of 2 or more cubes.

 x y z (Tet(x)  Cube(y)  Cube(z)  LeftOf(x,y)  LeftOf(x,z)  y  z)
2. The powerset P(S) of some set S, is composed of all subsets of set S (including the empty set and the set
S, itself). For example, for S = {1,2}, its powerset has 4 subsets, as it is P(S)= { {}, {1}, {2}, {1,2} }.
a) Indicate the powersets of the following sets Si:
S1.
{}
P(S1) = { { } }
S2.
{a}
P(S2) = { { }, {a} }
S3.
{1,2,3}
P(S3) = { { }, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} }
b) Prove by induction on the number of elements of set S, that the powerset P(S) of a set S with n elements is
composed of 2n distinct subsets of S. Sugestion: The powerset of a set with n+1 elements may be formed
from the powerset of a set with n elements, by considering for each of the subsets of the latter the inclusion
or not of the extra element.
Let us determine the cardinality of P(S), by induction on the number n of elements of S.
Base Step : n = 0  # P(S) = 20
For the set  with 0 elements, the empty set, its powerset P() is composed exclusively of itself, i.e.
P() = { }, thus having one element, which proves the base step, since # P() = 1 = 20 .
Induction Step: # P(S) = 2n  # P(S’) = 2n+1
With no loss of generality, let S have n elements, none of which is z. Then S’ = S  {z} has n+1
elements. By the induction hypothesis, P(S) has 2n subsets of S. But the powerset of S’ can be
constructed from all the subsets of S by adding to them, or not adding to them, element z. Hence, for
each subset in P(S) there are 2 subsets in P(S’), and the cardinality of P(S’) is twice that of P(S), i.e.
# P(S’) = 2 * #P(S) = 2* 2n = 2n+1,
which concludes the induction step.
Introduction to Logic
2005 / 2006 – 4th Test – 9 Jan 2006
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3. Show the wrong steps in the proof below, and draw a Tarski World that shows that the argument is not
valid. Note: Justifify your answer in the justification box, below.
1
x (Cube(x)  (Large(x)  FrontOf(x,a)))
2
x (Tet(x)  y (Cube(y)  FrontOf(x,y)))
3
4
xyz ((FrontOf(x,y)  FrontOf(y,z))  FrontOf(x,z))
Tet(b)
5
y (Cube(y)  FrontOf(b,y))
6
c Cube(c)  FrontOf(b,c)
Elim : 2, 4
Elim  : 6
7
Cube(c)
8
Cube(c)(Large(c)FrontOf(c,a)) Elim  : 1
9
FrontOf(c,a)
Elim : 7, 8
10
FrontOf(b,c)  FrontOf(c,a)
Intr  : 6 , 9
11
FrontOf(b,a)
Elim : 10, 3
12
a
Elim : 5, 6 - 11
FrontOf(b,a)
Tet(b)  FrontOf(b,a)
13
b
Intr : 4 - 12
Justification: In lines 5 e 11 the universal instantiation of the formulas 1 and 3, respectively was
omited. Such omission does not induce any problem regarding the validity of the proof. Neither thus the
elimination of predicate Cube(c) in line 10. The important error occurs in line 9, where the elimination
of  should lead (Large(c)  FrontOf(c,a)). From this formula one could not conclude FrontOf(c,a)
which is paramount to obtain the final formula, Tet(b)  FrontOf(b,a).
4. For the valid argument below, complete the proof made in the natural deduction Fitch system, filling in
the empty boxes.
1.
x ((Cube(x)  Tet(x)) 
y z Between(x,y,z))
2.
x (Cube(x)  Tet(x)  Dodec(x))
3.
4.
 x y z Between(z,x,y)
a
5.
 Dodec(a)
6.
Cube(a)  Tet(a)  Dodec(a)
Elim  : 2
7.
Cube(a)  Tet(a)
8.
9.
Cube(a)  Tet(a)
Dodec(a)
Reit
10.

Intr  : 5, 9
11.
Cube(a)  Tet(a)
Elim  : 10
: 7
12.
Cube(a)  Tet(a)
Elim  : 6, 7-8, 9-11
13.
(Cube(a) Tet(a))  yz Between(a,x,y)
Elim  : 1
14.
15.
yz Between(a,x,y)
b c Between(a,b,c))
Elim  : 12, 13
16.
xyz Between(z,x,y))
Intr  : 15
17.

Intr  : 3, 16
18.

Elim  : 14, 15 - 17
19.
Dodec(a)
Intr  : 5 - 17
20.
x Dodec(x)
Intr  : 4 - 19
Introduction to Logic
2005 / 2006 – 4th Test – 9 Jan 2006
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5. Use the Fitch system, to prove the validity of the argument below.
1.
x y ((Tet(x)  Cube(y))  Larger(x,y))
2.
 x y (Larger(x,y)  Smaller(x,y))
3.
Smaller (a,b)
4.
Tet(a)  Cube(b)
5.
(Tet(a)  Cube(b))  Larger(a,b)
Elim 
6.
Larger(a,b)
Elim  : 4,5

7.
Smaller(a,b)
8.
Larger(a,b)  Smaller(a,b)
Intr  : 6,7
9.
x y (Larger(x,y)  Smaller(x,y))
Intr  : 8

Intr  : 2,9
10.
11.
Smaller(a,b)
Intr  : 7 - 10
12.

Intr  : 3,11
13. (Tet(a)  Cube(b))
Intr  : 4 - 12
6. Prove the same argument (of problem 5) , but using this time the Resolution system.
a) Show which clauses must be contradictory.
1.Tet(x1)  Cube(y1)  Larger(x1,y1)
2.Larger(x2,y2)  Smaller(x2,y2)
3.Smaller(a,b)
4. Tet(a)
% Negation of Conclusion
5. Cube(b)
% Negation of Conclusion
b) Prove their unsatisfiability (by obtaining the empty clause through resolution steps in the clauses).
7. Cube(y1)  Larger(a,y1)
Res. 4, 1
{ x1/a }
8. Larger(a,b)
Res. 7, 5
{ y1/b }
Res. 8, 2
{ x2/a, y2/b }
9.
Smaller(a,b)
10.
Introduction to Logic
Res. 9, 3
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