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.