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