Show the steps required to put the following axioms into clause form:
∀x ∀y [On(x,y) ⇒ Above(x,y)]
∀x ∀y ∀z [Above(x,y) ∧ Above(y,z) ⇒ Above(x,z)]
Remove the implications:
∀x ∀y [¬On(x,y) ∨ Above(x,y)]
∀x ∀y ∀z [¬(Above(x,y) ∧ Above(y,z)) ∨ Above(x,z)]
Move negations down to the atomic level:
∀x ∀y [¬On(x,y) ∨ Above(x,y)]
∀x ∀y ∀z [¬Above(x,y) ∨ ¬Above(y,z) ∨ Above(x,z)]
Rename variables (We won't deduct points if you forgot to do this):
∀x ∀y [¬On(u,v) ∨ Above(u,v)]
∀x ∀y ∀z [¬Above(x,y) ∨ ¬Above(y,z) ∨ Above(x,z)]
Drop the existential quantifiers
¬On(u,v) ∨ Above(u,v)
¬Above(x,y) ∨ ¬Above(y,z) ∨ Above(x,z)
Example | Heuristic | Result |
1 | Initial model | Apple is red fragrant sphere with weight=4 |
2 | Drop link | Apple is red sphere with weight=4 |
3 | Forbid link | Apple is red sphere with weight=4, not inedible |
4 | Enlarge set | Apple is red or green sphere with weight=4, not inedible |
5 | Close interval | Apple is red or green sphere with 4≤weight≤7, not inedible |
6 | Climb tree | Apple is red or green rounded object with 4≤weight≤7, not inedible |
7 | none | Nothing learned, same model as above |
All the true successes involve not only George and Sally, but Frank as well. The false successes have nothing in common, except that George and Sally are involved. You should improve the rule to state that you only enjoy dinner parties attended by George, Sally, and Frank. Note that parties 5 and 7 have the same attendees, but different outcomes, so it is impossible to have a 100% accurate rule or rule set.