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above(a,d) rule (2) {X\a,Z\d} on(a,Y), above(Y,d) {Y\b} on(a,Y) - - PDF document
above(a,d) rule (2) {X\a,Z\d} on(a,Y), above(Y,d) {Y\b} on(a,Y) - - PDF document
above(a,d) rule (2) {X\a,Z\d} on(a,Y), above(Y,d) {Y\b} on(a,Y) above(b,d) answer: Y=b {X\b, Z\d} rule (2) on(b,Y), above(Y,d) {Y\c} above(c,d) on(b,Y) answer: Y=c rule (1) {X\c,Y\d} on(c,d) answer: yes All leaves are true, so the
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above(a,d) rule (2) {X\a,Z\d}
- n(a,Y), above(Y,d)
above(b,d) rule (2) {X\b, Z\d}
- n(b,Y), above(Y,d)
- n(a,Y)
{Y\c} {Y\b} above(c,d) rule (1) {X\c,Y\d}
- n(c,d)
answer: Y=b answer: yes
- n(b,Y)
answer: Y=c All leaves are true, so the root is true, i.e., above(a,d) is true.
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above rule (1) above, on above rule (1) above, on rule (1) above, on ...
This is a flaw in Prolog.
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