Deductive Logic, Warren Goldfarb, Hackett Publishing, 2003, Part IV, Exercise 5(c), p. 285

Show that (x)(y)(x = y ⊃ Gxy) is equivalent to (x)Gxx. The example is a good illustration of how The Convention works (I use a slightly modified version of The Convention, Symbolic Logic, Copi, 5th edition, Chpt 4.5, Quantification Rules):

The instantiating variable must be free, or the instantiating constant must occur, in the instantiating statement in at least those places where the general variable is free in the decapitated general statement.

A 'decapitated' statement is a statement where the quantifier has been taken off.

1. * (x)(y)(x = y ⊃ Gxy) ......... ACP
2. * (y)(x = y ⊃ Gxy) ......... 1 UI x/x
3. * x = x ⊃ Gxx ......... 2 UI y/x
4. * x = x ......... Id
5. * Gxx ......... 4,3 MP
6. * (x)Gxx ......... 5 UG
7. (x)(y)(x = y ⊃ Gxy) ⊃ (x)Gxx ......... 1-6 CP
8. * (x)Gxx ......... 8 ACP
9. * Gxx ......... 8 UI x/x
10. * (y)Gxy ......... 9 UG
11. * Gxy ......... 10 UI y/y
12. * Gxy ∨¬ (x = y) ......... 11 Add.
13. * ¬ (x = y) ∨ Gxy ......... 12 Comm.
14. * x = y ⊃ Gxy ......... 13 MI
15. (y)(x = y ⊃ Gxy) ......... 14 UG
16. (x)(y)(x = y ⊃ Gxy) ......... 15 UG
17. (x)Gxx ⊃ (x)(y)(x = y ⊃ Gxy) .........8-16 CP
18. [(x)Gxx ⊃ (x)(y)(x = y ⊃ Gxy)] • [(x)(y)(x = y ⊃ Gxy) ⊃ (x)Gxx] ......... 7,17 Conj.
19. (x)(y)(x = y ⊃ Gxy) ≡ (x)Gxx ......... 18 AB

Deductive Logic, Warren Goldfarb, Hackett Publishing, 2003, Part IV, Exercise 5(b), p. 285

The task: to show that the premises imply the conclusion.

1. (x)Gxx
2. (x)(y)[¬ (x = y) ⊃ (∃z)(Gxz • Gzy)]
3. ∴(x)(y)(∃z)(Gxz • Gzy)
4. * ¬ (x)(y)(∃z)(Gxz • Gzy) ......... IP
5. * (∃x)¬(y)(∃z)(Gxz • Gzy) ......... 4 QC
6. * (∃x)(∃y)¬(∃z)(Gxz • Gzy) ......... 5 QC
7. * (∃x)(∃y)(z)¬(Gxz • Gzy) ......... 6 QC
8. * (∃x)(∃y)(z)(¬Gxz ∨¬ Gzy) ......... 7 DeM.
9. * (∃x)(∃y)(z)(Gxz ⊃¬ Gzy) ......... 8 MI
10. * (∃y)(z)(Gaz ⊃¬ Gzy) ......... 9 EI x/a
11. * (z)(Gaz ⊃¬ Gzm) ......... 10 EI y/m
12. * Gaa ⊃¬ Gam ......... 11 UI z/a
13. * Gaa ......... 1 UI x/a
14. * ¬ Gam ......... 12,13 MP
15. * ¬ (a = m) ......... 13,14 Id
16. * (y)[¬ (a = y) ⊃ (∃z)(Gaz • Gzy)] ......... 2 UI x/a
17. * ¬ (a = m) ⊃ (∃z)(Gaz • Gzm) ......... 16 UI y/m
18. * (∃z)(Gaz • Gzm) ......... 15,17 MP
19. * Gar • Grm ......... 18 EI z/r
20. * Gar ......... 19 Simp.
21. * Grm ......... 19 Simp.
22. * Gar ⊃¬ Grm ......... 11 UI z/r
23. * ¬ Grm ......... 20,22 MP
24. * Grm • ¬ Grm ......... 21,23 Conj.
25. ¬ ¬ (x)(y)(∃z)(Gxz • Gzy) ......... 4-24 IP
26. (x)(y)(∃z)(Gxz • Gzy) ......... 25 DN

Deductive Logic, Warren Goldfarb, Hackett Publishing, 2003, Part IV, Exercise 5(a), p. 285

We are asked to show, using the laws of identity, that (∃x)Fxa and (x)¬ Fxb together imply ¬ (a = b). In the original example, a and b are respectively y and z, but I have used the first letters of the alphabet to indicate clearly that they are constants rather than unbound variables. We can proceed in a number of ways. I use indirect proof.

1. (∃x)Fxa
2. (x)¬ Fxb
3. ∴¬ (a = b)
4. * ¬ ¬ (a = b) ......... AIP
5. * a = b ......... 4 DN
6. * (∃x)Fxb ......... 1,5 Id
7. * Fmb ......... 6 EI x/m
8. * ¬ Fmb ......... 2 UI x/m
9. * Fmb • ¬ Fmb ......... 7,8 Conj.
10. ¬ ¬ ¬ (a = b) ......... 4-9 IP
11. ¬ (a = b) ......... 10 DN

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.4E, 1(k), p. 556

We have to derive the conclusion from the one premise given:

1. (x)[Fx ⊃(∃y)Gxy]
2. ∴(x)[Fx ⊃(∃y)(Gxy ∨ ¬ Hxy)]
3. * Fx ......... ACP
4. * Fx ⊃(∃y)Gxy ......... 1 UI x/x
5. * (∃y)Gxy ......... 3,4 MP
6. * (∃y)Gxy ∨ ¬ Hxy ......... 5 Add.
7. Fx ⊃(∃y)(Gxy ∨ ¬ Hxy) ......... 3-6 CP
8. (x)[Fx ⊃(∃y)(Gxy ∨ ¬ Hxy)] ......... 7 UG

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.4E, 1(a), p. 553

We have to derive: (x)Ax ≡ (x)(Ax • Ax). We start by making an assumption:

1. * (x)Ax ......... ACP
2. * Ax ......... 1 UI
3. * Ax ......... 2 Rep.
4. * Ax • Ax ......... 2,3 Conj.
5. * (x)(Ax • Ax) ......... 4 UG
6. (x)Ax ⊃(x)(Ax • Ax) ......... 1-5 CP
7. * (x)(Ax • Ax) ......... ACP
8. * Ax • Ax ......... 7 UI
9. * Ax ......... Simp.
10. * (x)Ax ......... 9 UG
11. (x)(Ax • Ax) ⊃(x)Ax ......... 7-10 CP
12. [(x)Ax ⊃(x)(Ax • Ax)] • [(x)(Ax • Ax) ⊃(x)Ax] ......... 6,11 Conj.
13. (x)Ax ≡ (x)(Ax • Ax) ......... 12 BE

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

I freely admit I have lost track of which problems I have covered across the many different books and which are still open. The really big ones that I know I haven't done yet would take up far too much time to first create in a Word document and then copy across to blogger. Anything over 30 lines is over an hour's job - more than I can afford to allocate to writing up this section. But, repetition in proving arguments, if indeed I have already done this one, is key to building up confidence. So, we show that the following is a theorem: (x)(y)(x = y ≡ y = x):

1. * x = y ......... ACP
2. * y = x ......... Id Comm.
3. x = y ⊃ y = x ......... 1-2 CP
4. * y = x ......... ACP
5. * x = y ......... Id Comm.
6. y = x ⊃x = y ......... 4-5 CP
7. (x = y ⊃ y = x) • (y = x ⊃x = y) ......... 3,6 Conj.
8. x = y ≡ y = x ......... 7 BE
9. (y)(x = y ≡ y = x) ......... 8 UG
10. (x)(y)(x = y ≡ y = x) ......... 9 UG

The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.6E, 1(c), p. 571

Show that [¬ (a = b) • b = c] ⊃ ¬ (a = c) is a theorem.

1. * ¬ (a = b) ......... ACP
2. * * b = c ......... ACP
3. * * ¬ (a = c)
4. * b = c ⊃ ¬ (a = c) ......... 2-3 CP
5. ¬ (a = b) ⊃ [b = c ⊃ ¬ (a = c)] ......... 1-4 CP
6. [¬ (a = b) • b = c] ⊃ ¬ (a = c) ......... 5 Exp.