Show that the statement ¬ (x)(∃y)[(Ax • Bx) ∨Cy] is equivalent to (∃x)(y)[ ¬ (Cy ∨Ax) ∨¬ (Cy ∨ Bx)]. The task involves primarily the use of DeMorgan's law and distribution. I show the equivalence one way. Of course, we need to show it both ways in order to be sure the statements are indeed equivalent. However, the reverse procedure is quite straightforward and involves practically working backwards through the steps 1-10.
- ¬ (x)(∃y)[(Ax • Bx) ∨Cy]
- (∃x)(y)[ ¬ (Ax • Bx) • ¬ Cy] ......... 1CQ
- (y)[ ¬ (Am • Bm) • ¬ Cy] ......... 2EI x/m
- ¬ (Am • Bm) • ¬ Cy ......... 3UI y/y
- (¬ Am ∨¬ Bm) • ¬ Cy ......... 4DeM
- (¬ Am • ¬ Cy ) ∨ (¬ Bm • ¬ Cy) ......... 5Dist.
- (¬ Cy • ¬ Am ) ∨ (¬ Cy • ¬ Bm) ......... 6Comm.
- ¬ (Cy ∨Am) ∨¬ (Cy ∨ Bm) ......... 7DeM
- (y)[ ¬ (Cy ∨Am) ∨¬ (Cy ∨ Bm)] ......... 8UG
- (∃x)(y)[ ¬ (Cy ∨Ax) ∨¬ (Cy ∨ Bx)]
Heard this yesterday http://www.bbc.co.uk/programmes/b00vcqcx and thought you also might enjoy listening to it :) - available for the next 6 days,
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a
Thanks for remembering about me and I am glad you listen to Malvyn Bragg's programme too. I'm a big fan. It was on in the other room when I was doing something else but it goes straight to my archive automatically so I will listen to it this weekend.
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