AnswerEX3-a: Difference between revisions
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Here the interpretation in predicate logic notation: | Here the interpretation in predicate logic notation: | ||
<nowiki>[[dog (walter) Ʌ enjoy-watching-soccer-together (lisa,tom)]] = FALSE | <nowiki>[[dog (walter) Ʌ enjoy-watching-soccer-together (lisa,tom)]] = FALSE <br/> | ||
because [[dog (walter)]]= TRUE | because [[dog (walter)]]= TRUE <br/> | ||
because I(walter)= walter and <walter> is an element of the set I(dog) | because I(walter)= walter and <walter> is an element of the set I(dog) <br/> | ||
but [[enjoy-watching-soccer-together (lisa,tom)]] = FALSE | but [[enjoy-watching-soccer-together (lisa,tom)]] = FALSE <br/> | ||
because I(lisa)= lisa, I(tom)= tom and <lisa,tom> is NOT an element of the set I(enjoy-watching-soccer-together). | because I(lisa)= lisa, I(tom)= tom and <lisa,tom> is NOT an element of the set I(enjoy-watching-soccer-together). <br/> | ||
====Navigation==== | ====Navigation==== | ||
*[[Interpretation_of_formulae_with_connectives|Link to the current exercise]] | *[[Interpretation_of_formulae_with_connectives|Link to the current exercise]] | ||
Revision as of 17:32, 27 January 2013
Here the interpretation in predicate logic notation:
<nowiki>dog (walter) Ʌ enjoy-watching-soccer-together (lisa,tom) = FALSE
because dog (walter)= TRUE
because I(walter)= walter and <walter> is an element of the set I(dog)
but enjoy-watching-soccer-together (lisa,tom) = FALSE
because I(lisa)= lisa, I(tom)= tom and <lisa,tom> is NOT an element of the set I(enjoy-watching-soccer-together).