Advanced Symbolic Logic
Homework set # 12 solutions
Page 189, # 5, by the Main Method
xy(Fxy  -Fyx)
x-Fxx
1. xy(Fxy  -Fyx)
2. -x-Fxx
Negate the conclusion
3. x--Fxx
quantifiere negation
4. –-Fxx
3 EI x/x
5. y(Fxy  -Fyx)
1, UI x/x
6. Fxx  -Fxx
5, UI x/y 3 and 6 are inconsistent
Page 189 # 6, by the Main Method
xyz(Fxy.Fyz.  -Fxz) x-Fxx
1. xyz(Fxy.Fyz.  -Fxz)
2. -x-Fxx
negation of the conclusion
3. x--Fxx
quantifier negation
4. --Fxx
3 EI x/x
5. yz(Fxy.Fyz.  -Fxz)
1, UI x/x
6. z(Fxx.Fxz.  -Fxz)
5,UI x/y
7. Fxx.Fxx.  -Fxx
6, UI x/z 3 and 7 are inconsistent
Page 198 #7, by the Main Method
xyz(Fxy.Fyz.  Fxz).x-Fxx xy(Fxy -Fyx)
1.xyz(Fxy.Fyz.  -Fxz)
2.x-Fxx
3.-xy(Fxy -Fyx)
negation of the conclusion
4.xy-(Fxy-Fyx)
quantifier negation
5.y-(Fxy-Fyx)
4. EI x/x
6.-(Fxy-Fyx)
5, EI y/y
7.-Fxx
2, UI x/x
8. yz(Fxy.Fyz.  Fxz) 1, UI x/x
9. z(Fxy.Fyz.  Fxz)
8, UI y/y
10.Fxy.Fyx.Fxx
9, UI x/z 6, 7, and 10 are inconsistent
Show that yx(Fx  Gy) and xy(Fx.-Gy) are inconsistent. We needn't negate either, just rip
off the quantifiers and demonstrate a truth functional inconsistency. The trick here is to recognize
that as the formulas stand, it is impossible to generate the necessary collision without violating a
restriction on EI, SO, purify the formulas and then re prenex!
1. yx(Fx  Gy)
8. yx(Fx.-Gy) 7, prenex
2. y(xFxGy) 1, purified
9. y(Fx  Gy)
4, EI x/x
3. xFxAyGy 2, purified
10. x(Fx.-Gy)
8, EI y/y
4. xy(FxGy) 3, prenex
11. FxGy
9, UI y/y
12. Fx.-Gy
10, UI x/x
5. xy(Fx.-Gy)
11 and 12 are inconsistent
6. x(Fx.y-Gy) 5, purified
7. xFx.y-Gy 6, purified
By either the method of pure existentials or the main method, show that xFxyxGx and y-Gy
together imply z-Fzy.
By the Main Method
1.xz(FxyGz)
2.y-Gy
3. zFzy
4. z(Fxy  Gz)
5. Fxy  Gz
6. –Gz
7. Fxy
z-Fzy
conclusion negated, QN
1 EI x/x
4 EI z/z
2 UI z/y
3 UI x/z 5-7 are inconsistent
Problem #4 p. 203
x(Fx.y(Gy.z(Fz.Hyz)..Hyx)), x(Gxy(Fy.Axy))  x(Fx.y(GyHyx))
1.
xyz(Fx:Gy.Fz.Hyz.Hyx) premise 1 prenex form
2.
xz(Gx.Fz.Hxz)
premise 2 prenex form, reletter
3.
xy(-Fx..Gy.-Hyx)
conclusion prenex form, negated, QN
4.
yz(Fx:Gy.Fz.Hyz.Hyx)
I EI x/x
5.
y(-Fx..Gy.-Hyx)
3 UI x/x
6.
-Fx..Gy.-Hyx
5 EI y/y
7.
z(Gy.Fz.Hyz)
2 UI y/x
8.
Gy.Fz.Hyz
7 EI z/z
9.
z(Fx:Gy.Fz.Hyz.Hyx)
4 UI y/y
10.
Fx:Gy.Fz.Hyz.Hyx
9 UI z/z 6,8 and 10 are inconsistent
Show, by the main method, that yzx(Fx.Gxy..Fy  Gzy) is valid.
1.-yzx(Fx.Gxy..Fy  Gzy)
formula negated
2. yzx(Fx.Gxy.-Fy.-Gzy)
QN
3. zx(Fx.Gxy.-Fy.-Gzy)
2 EI y/y
4. x(Fx.Gxy.-Fy.-Gzy)
3 UI z/z
5. Fw.Gwy.-Fy.-Gzy
4 EI w/x
6. x(Fx.Gxy.-Fy.-Gwy)
3 UI w/z
7. Fx.Gxy.-Fy.-Gwy
6 EI x/x 5 and 7 are inconsistnet