Translate the following into first order logic, then into clausal form
1. All birds have feathers.
2. Squigs have feathers
3. What do you get from resolving 1b with 2b.
4. Put BOTH these well-formed formulas (wffs) into clausal form: (Both means treat them as a pair of axioms about the world).
a) for all x, y [ On(x,y) implies Above(x,y)] (1)
b) for all x, y, z [ Above(x,y) & Above(y,z) implies Above(x,z)] (2)
5. Using the clause in 4 plus the statements:
Prove (by refutation resolution) that Above(B,Table).
Remember a refutation proof starts with the negation of the predicated to be proven, i.e. not Above(B,Table) (5)
I have numbered each of the clauses so that you explain each step in the proof by saying: resolving clause #n with clause #m. Number new clauses in your own proof.
To start you off:
Not Above(B,y) or not Above(y,Table) #6 by resolving #2 and #5.
There are several steps in the proof and you may need to search a while.