Lausearvutuse ja lihtsamate predikaatarvutuse ülesannete lahendamine - Redigeerimiste ajalugu

Tammet – 12. veebruar 2015, kell 13:35

2015-02-12T13:35:06Z

←Vanem redaktsioon		Redaktsioon: 12. veebruar 2015, kell 15:35
22. rida:		22. rida:
	Rewrite the formula as: p => p		Rewrite the formula as: p => p
	</pre>		</pre>

			NB! You may want to use http://logik.phl.univie.ac.at/~chris/gateway/formular-uk-zentral.html

	== [[Media:Geoffreasoningnotes.pdf\|Geoff notes]] page 19 first ==		== [[Media:Geoffreasoningnotes.pdf\|Geoff notes]] page 19 first ==

2013-01-31T19:32:00Z

Example

←Vanem redaktsioon		Redaktsioon: 31. jaanuar 2013, kell 21:32
117. rida:		117. rida:
	* exists X ( fast(main_competitor(main_competitor(X))) \| forall Y faster(X,Y) ) )		* exists X ( fast(main_competitor(main_competitor(X))) \| forall Y faster(X,Y) ) )

	== Example ==		== Example: first formula ==

	<pre>		<pre>

2013-01-31T19:31:20Z

Geoff notes page 22 nr 11 second

←Vanem redaktsioon		Redaktsioon: 31. jaanuar 2013, kell 21:31
116. rida:		116. rida:
	* forall X ( faster(main_competitor(X),laser) & fast(X) )		* forall X ( faster(main_competitor(X),laser) & fast(X) )
	* exists X ( fast(main_competitor(main_competitor(X))) \| forall Y faster(X,Y) ) )		* exists X ( fast(main_competitor(main_competitor(X))) \| forall Y faster(X,Y) ) )

			== Example ==

			<pre>
			faster(main_competitor(laser),commodore)
			faster(prelude,commodore)
			FALSE
			</pre>

2013-01-31T19:29:10Z

Pane tähele

←Vanem redaktsioon		Redaktsioon: 31. jaanuar 2013, kell 21:29
90. rida:		90. rida:
	Eeldame, et meil on siin võrduspredikaat olemas.		Eeldame, et meil on siin võrduspredikaat olemas.

	Näitena esimene ülesanne: forall X . Y. sum(X,Y)=sum(Y,X).		Näitena esimene ülesanne: forall X,Y. sum(X,Y)=sum(Y,X).

	Taustaks aritmeetika:		Taustaks aritmeetika:

2013-01-31T19:27:56Z

Geoff notes page 19 second

←Vanem redaktsioon		Redaktsioon: 31. jaanuar 2013, kell 21:27
58. rida:		58. rida:
	</pre>		</pre>

	Next, try to do at least one of these: ~~now~~ write conjectures for associativity of set intersection, and distribution of intersection over union.		Next, try to do at least one of these: write conjectures for associativity of set intersection, and distribution of intersection over union.

	=== Example and a question ===		=== Example and a question ===

2013-01-31T19:26:46Z

Geoff notes page 13/14

←Vanem redaktsioon		Redaktsioon: 31. jaanuar 2013, kell 21:26
20. rida:		20. rida:
	we see that p is equivalent to q.		we see that p is equivalent to q.
	Then (p \| q ) is equivalent to p and (p & q) is equivalent to p.		Then (p \| q ) is equivalent to p and (p & q) is equivalent to p.
	~~Rightmost part can thus be rewritten~~ as: p => p		Rewrite the formula as: p => p
	</pre>		</pre>

2013-01-31T19:26:21Z

Geoff notes page 13/14

←Vanem redaktsioon		Redaktsioon: 31. jaanuar 2013, kell 21:26
19. rida:		19. rida:
	((p => q) & (q => p) & (p \| q )) => (p & q)		((p => q) & (q => p) & (p \| q )) => (p & q)
	we see that p is equivalent to q.		we see that p is equivalent to q.
			Then (p \| q ) is equivalent to p and (p & q) is equivalent to p.
	Rightmost part can thus be rewritten as: p => p		Rightmost part can thus be rewritten as: p => p
	</pre>		</pre>

2013-01-31T19:25:48Z

Geoff notes page 13/14

←Vanem redaktsioon		Redaktsioon: 31. jaanuar 2013, kell 21:25
18. rida:		18. rida:
	((p => q) & (~q \| p) & (p \| q )) => (p & q)		((p => q) & (~q \| p) & (p \| q )) => (p & q)
	((p => q) & (q => p) & (p \| q )) => (p & q)		((p => q) & (q => p) & (p \| q )) => (p & q)
	p => p		we see that p is equivalent to q.
			Rightmost part can thus be rewritten as: p => p
	</pre>		</pre>

2013-01-31T19:22:39Z

←Vanem redaktsioon		Redaktsioon: 31. jaanuar 2013, kell 21:22
99. rida:		99. rida:
	forall X,Y +(s(X),Y) = s(+(X,Y)) .		forall X,Y +(s(X),Y) = s(+(X,Y)) .
	</pre>		</pre>

			== [[Media:Geoffreasoningnotes.pdf\|Geoff notes]] page 22 nr 11 second ==

			Given the 1 st order logic language:
			V = { V : V starts with uppercase }
			F = { holden/0, ford/0, honda/0 main_competitor/1 }
			P = { fast/1, faster/2 }
			and the interpretation:
			...

			Show the steps of the interpretation of:
			* faster(main_competitor(laser),commodore)
			* fast(main_competitor(main_competitor(prelude))
			* forall X ( faster(main_competitor(X),laser) & fast(X) )
			* exists X ( fast(main_competitor(main_competitor(X))) \| forall Y faster(X,Y) ) )

2013-01-31T19:14:36Z

Geoff notes page 19 second

←Vanem redaktsioon		Redaktsioon: 31. jaanuar 2013, kell 21:14
57. rida:		57. rida:

	Next, try to do at least one of these: now write conjectures for associativity of set intersection, and distribution of intersection over union.		Next, try to do at least one of these: now write conjectures for associativity of set intersection, and distribution of intersection over union.

			=== Example and a question ===

	As an example, the transitivity of subset: forall X, Y, Z (subset(X,Y) & subset(Y,Z) => subset(X,Z)).		As an example, the transitivity of subset: forall X, Y, Z (subset(X,Y) & subset(Y,Z) => subset(X,Z)).