AnswerEX3-a: Difference between revisions
Jump to navigation
Jump to search
No edit summary |
No edit summary |
||
| Line 5: | Line 5: | ||
because [[dog (walter)]]= TRUE <br/> | because [[dog (walter)]]= TRUE <br/> | ||
::because I(walter)= walter and <walter> is an element of the set I(dog) <br/> | |||
but [[enjoy-watching-soccer-together (lisa,tom)]] = FALSE <br/> | 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). <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 17:54, 27 January 2013
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 the set I(dog)
- 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).
- because I(lisa)= lisa, I(tom)= tom and <lisa,tom> is NOT an element of the set I(enjoy-watching-soccer-together).