In backward chaining, if it is desired to prove the conclusion C of a clause, the system tries to do so by proving the premises P1 ... Pn.
&forall x CAR(x) &and RED(x) &rarr EXPENSIVE(x)
Given this axiom, an attempt to prove that BMW1 is expensive would be reduced to the subproblems of proving that it is a car and that it is red.
Problems:
&forall x &forall y &forall z GREATER(x,y) &and GREATER(y,z) &rarr GREATER(x,z)
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