AnswerEX3-a: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 5: | Line 5: | ||
<nowiki>[[</nowiki>''dog (Walter) Ʌ enjoy-watching-soccer-together (Lisa,Tom)'']] = '''false''' <br/> | <nowiki>[[</nowiki>''dog (Walter) Ʌ enjoy-watching-soccer-together (Lisa,Tom)'']] = '''false''' <br/> | ||
because [[''dog (Walter'')]]= '''true''' <br/> | because <nowiki>[[</nowiki>''dog (Walter'')]]= '''true''' <br/> | ||
::because I(''Walter'')= <'''Walter'''> and <'''Walter'''> is an element of I(''dog'') <br/> | ::because I(''Walter'')= <'''Walter'''> and <'''Walter'''> is an element of I(''dog'') <br/> | ||
Revision as of 11:36, 28 January 2013
Sentence: Walter is a dog and lisa and tom enjoy watching soccer together.
Here the interpretation in predicate logic notation:
[[dog (Walter) Ʌ enjoy-watching-soccer-together (Lisa,Tom)]] = false
because [[dog (Walter)]]= true
- because I(Walter)= <Walter> and <Walter> is an element of I(dog)
- because I(Walter)= <Walter> and <Walter> is an element of I(dog)
but [[enjoy-watching-soccer-together (Lisa,Tom)]] = false
- because I(Lisa)= <Lisa>, I(Tom)= <Tom> and <Lisa,Tom> is NOT a set of I(enjoy-watching-soccer-together).
- because I(Lisa)= <Lisa>, I(Tom)= <Tom> and <Lisa,Tom> is NOT a set of I(enjoy-watching-soccer-together).
Conjunction (Ʌ): Both atomic formulae have to be true in order for the complex formula to be true.