Download more solutions

Short Answers
2.
If 1…2 are inconsistent predicate calculus sentences, then there is no
case in which they are all true relative to some interpretation. If so, then by
definition they imply anything (as there is no case in which they can be true and
the conclusion from them false), and so they imply also P & -P. But if a set of
sentences of the predicate calculus implies some other predicate calculus
sentence, then, by completeness theorem, there is a proof of that sentence from
the given set of sentences.
4.
In line (6) the dependency number ‘2’ is missing, which is what remains
when subtracting the dependency numbers of line (5) from that of line (3).
5.
Line (4) is not a contradiction that could be used with RAA. A
contradiction is something of the form ‘ & - ’, whereas line (4) has only the
form .
6.
Line (6) in the problem 4 is bad. No line is bad in the problem 5.
Proofs and counterexamples
3A:
1
2
1
1
1
1
1
8
9
9
8, 9
8
8
8
1, 8
1
1
18
19
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
(19)
(x)(y)[Qxy  (z)(Rzx  Rzy)]
-Qaa
(y)[Qay  (z)(Rza  Rzy)]
Qaa  (z)(Rza  Rza)
[Qaa  (z)(Rza  Rza)] & [(z)(Rza  Rza)  Qaa]
(z)(Rza  Rza)  Qaa
- (z)(Rza  Rza)
-(z)- (Rza  Rza)
-(Rba  Rba)
(z)-(Rza  Rza)
-(z)- (Rza  Rza) & (z)- (Rza  Rza)
--(Rba  Rba)
Rba  Rba
(z) (Rza  Rza)
(z)(Rza  Rza) & - (z)(Rza  Rza)
--(z)- (Rza  Rza)
(z)- (Rza  Rza)
- (Rba  Rba)
Rba
A
A
1 UE
2 UE
4 Def.
5 &E
2, 6 MTT
A
A
9 EI
8, 10 &I
9,11 RAA
12 DN
13 UI
7, 14 &I
8, 15 RAA
16 DN
A
A
18
18
18
1
1
(20)
(21)
(22)
(23)
(24)
(25)
Rba  Rba
- (Rba  Rba) & (Rba  Rba)
--Qaa
Qaa
Qaa
(x)Qxx
19, 19 CP
18, 20 &I
2, 18 RAA
22 DN
17,18,23 EE
24 UI
3B:
X = {1, 2}
Ext(Qxy) = {<1,2>,<1,1>,<2,2>}
Ext(Rxy) = {<2,2>}
3Cii:
X={1, 2, 3}
Ext(Rxy)=
Ext(Qxy)={<1,1>, <1,2>,<1,3>,<2,2>, <2,1>, <2,3>,<3,3>,<3,1>,<3,2>}