The Logic Book, M. Bergmann, J. Moor, J. Nelson, McGraw Hill, 2004, 10.4E, 5(a), p. 557

We are asked to show that the two: (x)Ax and (x)(Ax • Ax) are equivalent. In other words, we must be able to go from one to the other and back.

1. * (x)Ax ......... ACP
2. * Ax ......... 1 UI x/x
3. * Ax ......... 2 Rep.
4. * Ax • Ax ......... 2,3 Conj.
5. * (x)(Ax • Ax) ......... 4 UG
6. (x)Ax ⊃ (x)(Ax • Ax) ......... 1-6 CP

1. * (x)(Ax • Ax) ......... ACP
2. * Ax • Ax ......... 1 UI x/x
3. * Ax ......... 2 Simp.
4. * (x)Ax ......... 3 UG
5. (x)(Ax • Ax) ⊃ (x)Ax ......... 1-4 CP

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