Wednesday, 27 April 2011

Predicate Logic, Howard Pospesel, Prentice Hall, 2003, Chpt. 14, problem 25, p. 213

The argument:

"A set A is a subset of a set B iff there is no member of A that is not a member of B. The empty set E is a subset of every set, since for any set C there is no member of E that is not a member of C, simply because there is no member of E."

The proof:
  1. (x)(Ax ⊃Bx) ≡ ¬ (∃x)(Ax • ¬ Bx)
  2. ¬ (∃x)(Ex • ¬ Cx)
  3. ¬ (∃x)Ex
  4. ∴(x)[Ex ⊃(Ax • Bx • Cx)]
  5. * Ex ......... ACP
  6. * (x) ¬ Ex ......... 3 QC
  7. * ¬ Ex ......... 6 UI
  8. * ¬ Ex ∨(Ax • Bx • Cx) ......... 7 Add.
  9. * Ax • Bx • Cx ......... 8,5 DS
  10. Ex ⊃(Ax • Bx • Cx) ......... 5-9 CP
  11. (x)[Ex ⊃(Ax • Bx • Cx)] ......... 10 UG

Thursday, 21 April 2011

Symbolic Logic, D. Jacquette, Wadsworth, 2001, Chpt. 8, IV(13), p.435

The argument:

"All treaties with the Lakota nation are to be honoured. Thus, whatever is not to be honoured is not a treaty with the Lakota nation, and everything is either not a treaty with the Lakota nation or it is to be honoured."

The simplest form of the proof requires no more than working with the definitions as the conclusion is merely a restatement of the premise. Contraposition and material implication have been used.
  1. (x)(Txl ⊃Hx)
  2. (x)(¬ Hx ⊃¬ Txl) • (x)(¬ Txl ∨Hx)
  3. (x)(¬ Hx ⊃¬ Txl) ......... 1 Contrap.
  4. (x)(¬ Txl ∨Hx) ......... 1 MI
  5. (x)(¬ Hx ⊃¬ Txl) • (x)(¬ Txl ∨Hx) ......... 3,4 Conj.

Thursday, 14 April 2011

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.6E, 4(b), p. 572

The argument:

"Hyde killed some innocent person. But Jekyll is Hyde. Jekyll is a doctor. Hence, some doctor killed some innocent person."

The proof, with the universe of discourse being 'persons':
  1. (∃x)(Ix • Khx)
  2. j = h
  3. Dj
  4. ∴(∃x)[Dx • (∃y)(Iy • Kxy)]
  5. Im • Khm ......... 1 EI x/m
  6. Khm ......... 5 Simp.
  7. Kjm ......... 3,6 Id
  8. Im ......... 5 Simp.
  9. Im • Kjm ......... 7,8 Conj.
  10. (∃y)(Iy • Kjy) ......... 9 EG
  11. Dj (∃y)(Iy • Kjy) ......... 3,10 Conj.
  12. (∃x)[Dx • (∃y)(Iy • Kxy)] ......... 11 EG

Thursday, 7 April 2011

The 'undisciplined man' argument

In Plato’s Gorgias, Socrates expounds on the virtues of temperance, discipline and order to Callicles. At one point, the argument proceeds along these lines:

This I consider to be the mark to which a man should look throughout his life, and all his own endeavors and those of his city he should devote to the single purpose of so acting that justice and temperance shall dwell in him who is to be truly blessed. He should not suffer his appetites to be undisciplined and endeavor to satisfy them by leading the life of a brigand—a mischief without end. For such a man could be dear neither to any other man nor to God, since he is incapable of fellowship, and where there is no fellowship, friendship cannot be. Wise men, Callicles, say that the heavens and the earth, gods and men, are bound together by fellowship and friendship, and order and temperance and justice, and for this reason they call the sum of things the 'ordered' universe, my friend, not the world of disorder or riot.

From this, we winkle out:

For such a man could be dear neither to any other man nor to God, since he is incapable of fellowship, and where there is no fellowship, friendship cannot be.

and see if the argument really stands up logically. I first saw the challenge in Howard Pospesel’s Predicate Logic (2003), where we are set the task of translating the argument into the language of first order logic and then proving it.

The challenge lives up to its designation. First, because the tools we are given by Professor Pospesel are, in my view, somewhat unhelpful. Second, because, again, in my view, the argument is an enthymeme, i.e., it is missing a premise.

We are instructed to follow this dictionary: domain: persons; Ux = x is undisciplined; Hx = x is human; Rxy = x can have friendship with y; g = God; Exy = x can have fellowship with y. The argument is to consist of two premises.

The reason I find this unhelpful is, first, that the universe of discourse is set at ‘persons’, while, and here I invite correction, ‘any other man’ is introduced by the predicate letter Hx, where x is human. The overlap is clearly substantial. Second, the relation ‘is dear to’ does not appear in the dictionary, except presumably as ‘x can have friendship with’. This is not a problem as long as we know that this reading is intended (otherwise the argument will not work).

The bigger problem is the elliptical nature of the argument. As it stands, it falls. Or else I’m missing something. So here is the formulation I propose using the classical argument structure. The third premise has been inserted.

  1. An undisciplined man is not a fellow of any other man.
  2. If a man is not a fellow of any other, then he is not their friend either.
  3. If a man is not a friend of any other man, then he is not a friend of God.
  4. Therefore, an undisciplined man is neither a friend of any other man nor a friend of God.

As for the predicate ‘x is human’, I prefer to use identity – it organizes our universe of discourse more effectively. Following the original dictionary in all other respects: Rxy = x can have friendship with y; g = God; Exy = x can have fellowship with y, we get:

  1. (x)(y){[Ux • ¬ (x = y)] ⊃¬ Exy}
  2. (x)(y){[¬ (x = y) • ¬ Exy] ⊃¬ Rxy}
  3. (x){(y)[¬ (x = y) ⊃¬ Rxy] ⊃¬ Rxg}
  4. (x){Ux ⊃ ¬ {(∃y)[ ¬ (x = y) • Rxy] ∨ Rxg}}


The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.6E, 3(d), p. 572

Show that the following is a theorem: (x)(y)(z)[(x = y • y = z)⊃ x = z]. The statement illustrates the transitive property of relations. The proof is quite straightforward:

(x)(y)(z)[(x = y • y = z)⊃ x = z]
  1. * x = y • y = z ......... ACP
  2. * x = y ......... 1 Simp.
  3. * y = z ......... 1 Simp.
  4. * x = z ......... 2,3 Id
  5. (x = y • y = z)⊃ x = z ......... 1-4 CP
  6. (z)[(x = y • y = z)⊃ x = z] ......... 5 UG
  7. (y)(z)[(x = y • y = z)⊃ x = z] ......... 6 UG
  8. (x)(y)(z)[(x = y • y = z)⊃ x = z] ......... 7 UG