12/26/06 under construction
You now have to tools to appraise arguments to the level of detail offered by predicate logic.
Let us run through how these might be used with a long and difficult example.
Consider the argument
Anyone on the committee who knew the nominee would vote for the nominee, if free to do so. Everyone on the committee was free to vote for the nominee except those who were either instructed not to do so by the party caucus or who had pledge support to someone else. Everyone on the committee knew the nominee. No one who knew the nominee had pledged support to anyone else. Someone on the committee did not vote for the nominee. So the party caucus had instructed some members of the committee not to vote for the nominee. [This example is adapted from Copi]
First we symbolize it
(x) ((Cx.Kx.Fx)⊃Vx )
(x) (Cx⊃(Fx∨Ix∨Px))
(x) (Cx⊃Kx)
∼(∃x)(Kx.Px)
(∃x)(Cx.∼Vx)
∴
(∃x)(Cx.Ix)Cx= x is on the committee
Fx= x is free to vote for the nominee
Ix= x has been instructed by the party caucus not to vote for the nominee
Kx= x knew the nominee
Px= x had pledged support to someone else
Vx= x votes for the nominee
Then we should have a quick look at the semantics. We start by producing an interpretation in which the conclusion is false, say
Now the first four premises are true in this, but the fifth one is not, we need for (∃x)(Cx.∼Vx) to be true (ie for there to be an x which is C but not V). We try
Now, the conclusion is true, and so too is premise 5, but premises 2 and 3, (x)(Cx⊃(Fx∨Ix∨Px)) and (x)(Cx⊃Kx), are not true.
We need to make our a have the property K to get (x)(Cx⊃Kx) to be true, and it must also have one of the properties F, I, or P to get premise 2 to be true. This is all starting to get a little sticky.
At this point we should wonder whether the argument might be valid, and whether we can produce a derivation of it. The derivation starts
1. (x)((Cx.(Kx.Fx))⊃Vx) 2. (x)(Cx⊃(Fx∨(Ix∨Px))) 3. (x)(Cx⊃Kx) 4. ∼(∃x)(Kx.Px) 5. (∃x)(Cx.∼Vx) 6. ? ? << 7. (∃x)(Cx.Ix) ?
The first point to note is that even if we find a proof there may be other better, perhaps shorter and more elegant, proofs. A preliminary analysis picks out three things: a) the conclusion may well have been obtained by EG, so really we are looking for something like (Ca.Ia) as the penultimate line, b) there is only one existentially quantified formula there, line 5, and that should be instantiated early in the piece (when you have a mixture of universally and existentially quantified formulas, usually it is best to instantiate the existential ones first, then plug the instantiating constants as instantiations into the universal quantifiers, c) the negative existential of line 4 is unusable as it stands, it needs to be converted to a universally quantified formula.
1. (x)((Cx.(Kx.Fx))⊃Vx) 2. (x)(Cx⊃(Fx∨(Ix∨Px))) 3. (x)(Cx⊃Kx) 4. ∼(∃x)(Kx.Px) 5. (∃x)(Cx.∼Vx) 6. (x)∼(Kx.Px) 4 QN 7. (x)(∼Kx∨∼Px) 6 De M 8. (x)(Kx⊃∼Px) 7 Impl 9. Cu.∼Vu 5 EI 10. ? ? << 11. (∃x)(Cx.Ix) ?
At this point the existential quantifier of line 5 has been instantiated, using the term 'u', this leads us to expect that the penultimate line will be Cu.Iu (and indeed we can get Cu from line 9. The negative existential of line 4 has been transformed. Now, there is a matter of personal taste here, I reason better with 'if... thens' and Modus Ponens, rather than, in general, Disjunctive Syllogism so I have transformed it to an 'if.. then' in line 8. What we are looking for now is Iu (with that the proof is finished). Now, the only place at all the I appears is in line 2, so we will proceed from there.
1. (x)((Cx.(Kx.Fx))⊃Vx) 2. (x)(Cx⊃(Fx∨(Ix∨Px))) 3. (x)(Cx⊃Kx) 4. ∼(∃x)(Kx.Px) 5. (∃x)(Cx.∼Vx) 6. (x)∼(Kx.Px) 4 QN 7. (x)(∼Kx∨∼Px) 6 De M 8. (x)(Kx⊃∼Px) 7 Impl 9. Cu.∼Vu 5 EI 10. Cu⊃(Fu∨(Iu∨Pu)) 2 UI 11. Cu 9 Simp. 12. Fu∨(Iu∨Pu) 10,11 M.P. 13. ? ? << 14. (∃x)(Cx.Ix) ?
At this point, we are looking for Iu, and we have Fu∨(Iu∨Pu) as line 12. What might be a possibility here is to get ~Fu and ~Pu and that would do it. F only appears in premise 1 and 2 and we have already used premise 2. This might be the time to use premise 1. We are trying to get, as our next intermediate goal ~Fu.
1. (x)((Cx.(Kx.Fx))⊃Vx) 2. (x)(Cx⊃(Fx∨(Ix∨Px))) 3. (x)(Cx⊃Kx) 4. ∼(∃x)(Kx.Px) 5. (∃x)(Cx.∼Vx) 6. (x)∼(Kx.Px) 4 QN 7. (x)(∼Kx∨∼Px) 6 De M 8. (x)(Kx⊃∼Px) 7 Impl 9. Cu.∼Vu 5 EI 10. Cu⊃(Fu∨(Iu∨Pu)) 2 UI 11. Cu 9 Simp. 12. Fu∨(Iu∨Pu) 10,11 M.P. 13. (Cu.(Ku.Fu))⊃Vu 1 UI 14. ∼Vu.Cu 9 Com 15. ∼Vu 14 Simp. 16. ∼(Cu.(Ku.Fu)) 13,15 M.T. 17. ∼Cu∨∼(Ku.Fu) 16 De M 18. ? ? << 19. (∃x)(Cx.Ix) ?
This is proceding, we are still looking for ~Fu.
1. (x)((Cx.(Kx.Fx))⊃Vx) 2. (x)(Cx⊃(Fx∨(Ix∨Px))) 3. (x)(Cx⊃Kx) 4. ∼(∃x)(Kx.Px) 5. (∃x)(Cx.∼Vx) 6. (x)∼(Kx.Px) 4 QN 7. (x)(∼Kx∨∼Px) 6 De M 8. (x)(Kx⊃∼Px) 7 Impl 9. Cu.∼Vu 5 EI 10. Cu⊃(Fu∨(Iu∨Pu)) 2 UI 11. Cu 9 Simp. 12. Fu∨(Iu∨Pu) 10,11 M.P. 13. (Cu.(Ku.Fu))⊃Vu 1 UI 14. ∼Vu.Cu 9 Com 15. ∼Vu 14 Simp. 16. ∼(Cu.(Ku.Fu)) 13,15 M.T. 17. ∼Cu∨∼(Ku.Fu) 16 De M 18. ∼∼Cu 11 DN 19. ∼(Ku.Fu) 17,18 D.S. 20. ∼Ku∨∼Fu 19 De M 21. Cu⊃Ku 3 UI 22. Ku 21,11 M.P. 23. Ku⊃∼Fu 20 Impl 24. ∼Fu 23,22 M.P. 25. ? ? << 26. (∃x)(Cx.Ix) ?
Now we have it, and we can use it with line 12.
1. (x)((Cx.(Kx.Fx))⊃Vx) 2. (x)(Cx⊃(Fx∨(Ix∨Px))) 3. (x)(Cx⊃Kx) 4. ∼(∃x)(Kx.Px) 5. (∃x)(Cx.∼Vx) 6. (x)∼(Kx.Px) 4 QN 7. (x)(∼Kx∨∼Px) 6 De M 8. (x)(Kx⊃∼Px) 7 Impl 9. Cu.∼Vu 5 EI 10. Cu⊃(Fu∨(Iu∨Pu)) 2 UI 11. Cu 9 Simp. 12. Fu∨(Iu∨Pu) 10,11 M.P. 13. (Cu.(Ku.Fu))⊃Vu 1 UI 14. ∼Vu.Cu 9 Com 15. ∼Vu 14 Simp. 16. ∼(Cu.(Ku.Fu)) 13,15 M.T. 17. ∼Cu∨∼(Ku.Fu) 16 De M 18. ∼∼Cu 11 DN 19. ∼(Ku.Fu) 17,18 D.S. 20. ∼Ku∨∼Fu 19 De M 21. Cu⊃Ku 3 UI 22. Ku 21,11 M.P. 23. Ku⊃∼Fu 20 Impl 24. ∼Fu 23,22 M.P. 25. Iu∨Pu 12,24 D.S. 26. Ku⊃∼Pu 8 UI 27. ∼Pu 26,22 M.P. 28. ? ? << 29. (∃x)(Cx.Ix) ?
Now we are looking really good, we are just a step away from Iu and having the proof finished.
1. (x)((Cx.(Kx.Fx))⊃Vx) 2. (x)(Cx⊃(Fx∨(Ix∨Px))) 3. (x)(Cx⊃Kx) 4. ∼(∃x)(Kx.Px) 5. (∃x)(Cx.∼Vx) 6. (x)∼(Kx.Px) 4 QN 7. (x)(∼Kx∨∼Px) 6 De M 8. (x)(Kx⊃∼Px) 7 Impl 9. Cu.∼Vu 5 EI 10. Cu⊃(Fu∨(Iu∨Pu)) 2 UI 11. Cu 9 Simp. 12. Fu∨(Iu∨Pu) 10,11 M.P. 13. (Cu.(Ku.Fu))⊃Vu 1 UI 14. ∼Vu.Cu 9 Com 15. ∼Vu 14 Simp. 16. ∼(Cu.(Ku.Fu)) 13,15 M.T. 17. ∼Cu∨∼(Ku.Fu) 16 De M 18. ∼∼Cu 11 DN 19. ∼(Ku.Fu) 17,18 D.S. 20. ∼Ku∨∼Fu 19 De M 21. Cu⊃Ku 3 UI 22. Ku 21,11 M.P. 23. Ku⊃∼Fu 20 Impl 24. ∼Fu 23,22 M.P. 25. Iu∨Pu 12,24 D.S. 26. Ku⊃∼Pu 8 UI 27. ∼Pu 26,22 M.P. 28. Pu∨Iu 25 Com 29. Iu 28,27 D.S. 30. Cu.Iu 11,29 Conj. 31. (∃x)(Cx.Ix) 30 EG
The example argument is valid. The derivation that proves that it is valid is long and difficult. Also, derivations are not unique. There are can be many different proofs of a theorem (or that an argument is valid). There may even be an 8 line proof of the above!
Most experienced logicians are pretty familiar with the following theorems. Although the theorems are stated in terms of Fx and Gx etc. most of them hold of most formulas. Derive them.
Proofs
a) ∴ ((x)F)≡F (*this is true for any formula F provided that x is not free in F*)
b)∴ ((x)Fx)≡((y)Fy) (*change of variables*)
c) ∴ ((x)(y)Fxy)≡((y)(x)Fxy) (*permuting universal quantifiers*)
d) ∴ ((∃x)(∃y)Fxy)≡((∃y)(∃x)Fxy) (*permuting existential quantifiers*)
e) ∴ ((∃x)(y)Fxy)⊃((y)(∃x)Fxy)
f) Note that the following is not a theorem, see if you can produce a counter example
∴((y)(∃x)Fxy)⊃((∃x)(y)Fxy)
(*this is the 'everything has a cause therefore some (one) thing is the cause of everything' fallacy*)
Proofs
g) ∴ ((∃x)Fx)≡(∼(x)∼Fx) (*equivalence of connectives*)
h) ∴ (∼(∃x)∼Fx)≡((x)Fx) (*equivalence of connectives*)
i) ∴ (x)(Fx≡Gx)⊃((x)Fx≡(x)Gx)
(*but j) note that (or prove that) ∴(x)(Fx≡Gx)≡((x)Fx≡(x)Gx) is invalid*)
k) ∴ (x)(Fx≡Gx)⊃((∃x)Fx≡(∃x)Gx)
l) ∴ (x)(Fx⊃Gx)⊃((x)Fx⊃(x)Gx)
Proofs
m) ∴ (x)(Fx⊃Gx)⊃((∃x)Fx⊃(∃x)Gx)
n) ∴ ((x)Fx⊃(x)Gx)⊃(∃x)(Fx⊃Gx)
o) ∴ (x)(Fx.Gx)≡((x)Fx.(x)Gx)
p) ∴ (x)(Fx∨Gx)⊃((x)Fx∨(∃x)Gx)
q) ∴ ((x)Fx∨(x)Gx)⊃(x)(Fx∨Gx)
r) ∴ ((∃x)Fx∨(∃x)Gx)≡(∃x)(Fx∨Gx)
Interpretations Applet
