Prolog programming language and proof trees -

Prolog programming language and proof trees -

recall proof meta-circular

solve(true, true). solve([], []). solve([a|b],[proofa|proofb]) :- solve(a,proofa), solve(b, proofb). solve(a, node(a,proof)) :- rule(a,b), solve(b,proof).

assume 3rd rule of interpreter altered, while other rules of interpreter unchanged, follows:

% signature: solve(exp, proof)/2 solve(true, true). solve([], []). solve([a|b], [proofa|proofb]) :- solve(b, proofb), %3 solve(a, proofa). solve(a, node(a, proof)) :- rule(a, b), solve(b, proof).

consider proof tree created query in both versions. can variable substitution achieved in 1 version only? explain. can true leaf move other side of left infinite branch? explain. in both questions give illustration if reply positive. how influence on proof?

please help me ! tx

(i have lot of reservations against meta-interpreter. first reply question had)

in meta-interpreter reifying (~ implementing) conjunction. , implement prolog's conjunction. have 2 different versions how interpret conjunction. 1 time prove first, b. opposite.

think of

p :- p, false.

and

p :- false, p.

the sec version produce finite failure branch, whereas first produce infinite failure branch. effect of using 1 or other meta-interpreter. note "error" might 1 time again mitigated interpreting meta-interpreter itself!

see this answer might clarify notions bit.

there other ways implement conjunction (via binarization) ; such next level of meta-interpreter no longer able compensate.

finally comment on style of meta-interpreter: mixing lists , other terms. in fact [true|true] true. avoid such representation means. either stick traditional "vanilla" representation operates on syntax tree of prolog rules. is, conjunction represented (',')/2. or stick lists. not mix lists , other representations.

prolog