Quantifiers - University of Nebraska Omaha
Quantifiers - University of Nebraska Omaha
Quantifiers - University of Nebraska Omaha
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
MATH 2230 — FALL 2013 19<br />
Now, suppose that x 0 is arbitrary but fixed. It follows from (⇤) that<br />
(x 0 )istrue,andby(⇤⇤) wegetthat (x 0 )holds.Therefore,<br />
(x 0 ) ^ (x 0 )<br />
is true. Since x 0 was arbitrary, we conclude that (8x)( (x) ^ (x)) is<br />
true.<br />
Consequently, the desired implication is true and the pro<strong>of</strong> <strong>of</strong> the<br />
biconditional is complete.<br />
(b) To show this biconditional we will prove the two implications<br />
separately.<br />
()) Assume (9x)( (x) _ (x)) holds true. Let x 0 be such that<br />
(x 0 ) _ (x 0 )<br />
is true. Hence, one <strong>of</strong> the statements,<br />
consider the two possible cases.<br />
(x 0 )or (x 0 ), is true. Let us<br />
Case 1: (x 0 )istrue.<br />
Then, x 0 witnesses that (9x) (x) holds true. Consequently, the disjunction<br />
(9x) (x) _ (9x) (x)<br />
holds true.<br />
Case 2: (x 0 )istrue.<br />
Then, similarly as in the previous case, except interchanging and ,<br />
we conclude that the disjunction<br />
holds true.<br />
(9x) (x) _ (9x) (x)<br />
In both cases we concluded that the disjunction (9x) (x)_(9x) (x)<br />
holds true. Therefore the implication is justified.<br />
(() Assume (9x) (x) _ (9x) (x) holdstrue. Thus,either(9x) (x)<br />
or (9x) (x) istrueandthisleadstotwopossiblecases.<br />
Case 1: (9x) (x) istrue.<br />
Then we may choose an x 0 so that (x 0 )holds. Itthenfollowsthat<br />
(x 0 ) _ (x 0 )istrue.Consequently,x 0 witnesses that<br />
(9x)( (x) _ (x))<br />
holds true.<br />
Case 2: (9x) (x) istrue.<br />
By interchanging and in the previous case, we conclude that<br />
(9x)( (x) _ (x))
Change language