Advanced Symbolic Logic
Homework set # 13 solutions
(1)
Falsify ∃x∀y∃z(Fyz → Fxz. → .Fxx → Fyx). Let Fxy = x is identical with y and limit
the universe to the positive integers. Using material implication and limiting the z quantifier to
the antecedent of the conditional, this formula says "Either there is an integer that is not identical
with itself, or every integer is identical to it, or there is an integer identical to every integer to
which the first identified integer is not identical." Plainly false. Remember, if a formula has a
true interpretation it has one in the universe of natural numbers.
Falsification is satisfaction of the negation, so if we can satisfy
-∃x∀y(z(Fyz → Fxz) → .Fxx → Fyx) we are done. Driving the z, which is confined to the
antecedent, in make the procedure clearer. Thus, we are trying to satisfy xy(z(Fyz → Fxz).
Fxx. -Fyx). If F = ≤ and our universe is the universe of natural numbers, then our claim is true,
thus the original is falsified.
(2) Show that "Everybody loves a lover," on its most likely construal, together with "Bob loves
Carol" as a further premise implies that "Ted loves Alice." You may use either the main method
or the method of pure existentials (Boolos).
1. xyz(Lxy  Lzx) Lda
2. Lbc
3. –Lda
4. yz(Lby  Lzb) 1 UI b/x
5. z(Lbc  Lzb)
4 UI c/y
6. LbcLab
5 UI a/z
7. yz(Lay  Lza) 1 UI a/x
8. z(Lab  Lza)
7 UI b/y
9. Lab  Lda
9 UI a/z 2,3,6, and 10 are inconsistent
If we use Pure existentials, we show the validity of xyz(Lxy  Lzx. Lbc.  Lda).
3existentially quantified variables and 4 free variables produces a 64 way alternation that I shall
not produce here. 2 of theose alterns, LbcLab.Lbc.Lta : v : LabLta.Lbc.Lta, however
are enough to show validity.
(3) Use the main method, plus rules of passage (9) and (10) but no others, to show that
xuyv(Gy  Fx.Fu  Gv) implies yx(Fx  Gy).
1. ∃x∃u∀y∀v(Gy → Fx.Fu → Gv)
∀y∃x(Fx ↔ Gy)
2. -∀y∃x(Fx ↔ Gy)
3. yx-(Fx  Gy)
4. ∃u∀y∀v(Gy → Fx.Fu → Gv)
1 EI x/x
5. ∀y∀v(Gy → Fx.Fu → Gv)
4 EI u/u
6. x-(Fx  Gy)
3 EI y/y
7. -(Fx  Gy)
6 UI x/x
8. -(Fu  Gy)
6 UI u/x
9. ∀v(Gy → Fx.Fu → Gv)
5 UI y/y
10. Gy → Fx.Fu → Gy
9 UI y/u 7,8, and 10 are inconsistent
(4)
Show that "Everybody loves my baby but my baby loves nobody but me" implies "I am
my baby." To simplify things a little, you may symbolize the conclusion as if it read "My baby
is me." (Cartwright)
1. xLxb
2. x(Lbx  x =m)
m=b
3. m  b
neg of conclusion
4. Lbb
1 UI b/x
5. Lbb  b = m
2 UI b/x
6. b = b
Axiom II
7. b = b. b =m.  m = b
Axiom I 3-7 are inconsistent
(5)
Use the main method, plus the axioms of identity, to show that xFx and y-Fy imply
xy(x≠y).
1. xFx
∀x∃y(x≠y)
2. y-Fy
3. -∀x∃y(x≠y)
4. xy(x=y)
5. Fx
1 EI x/x
6. –Fy
2 EI y/y
7. y(z = y)
4 EI z/x
8. z = x
7 UI x/y
9. y=x
7 UI y/y
10. Fz. z = y. Fy
Axiom I
11. –Fz. z=x. -Fx
Axiom I 5,6,8.9.10, and 11 are inconsistent
(6)
Show that "There is a dodo who admires every dodo," "Dodos all admire whatever
admire them," and "A dodo who admires any dodo admires every dodo," imply "Each dodo
admires every dodo." (Koopman)
1.
xy(Dx.Dy  Axy)
2.
xy(Dx . Ayx  Axy)
3.
xyz(Dx.Dy.Axy..DzAxz)
4.
xy(Dx.Dy.-Axy)
negation of conclusion
5.
y(Ds.Dy.-Asy)
4, EI s/x
6.
Ds.Dt.-Ast
5, EI t/y
7.
y(Dw.Dy  Awy)
1, EI w/x
8.
Dw.Ds  Aws
7, UI s/y
9.
y(Ds . Ays  Asy)
2, UI s/x
10.
Ds . Aws  Asw
9, UI w/y
11.
yz(Ds.Dy.Asy..DzAsz) 3, UI s/t
12.
z(Ds.Dw.Asw..DzAsz)
11, UI w/y
13.
Ds.Dw.Asw..DtAst
12, UI t/z
6,8,10 &13 are inconsistent