(0) Obligation:

Clauses:

gopher(nil, nil).
gopher(cons(nil, Y), cons(nil, Y)).
gopher(cons(cons(U, V), W), X) :- gopher(cons(U, cons(V, W)), X).

Query: gopher(g,a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

(2) Obligation:

Clauses:

gopherA(nil, nil).
gopherA(cons(nil, T4), cons(nil, T4)).
gopherA(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23))).
gopherA(cons(cons(cons(T34, T35), T36), T37), T39) :- gopherA(cons(T34, cons(T35, cons(T36, T37))), T39).

Query: gopherA(g,a)

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
gopherA_in: (b,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

gopherA_in_ga(nil, nil) → gopherA_out_ga(nil, nil)
gopherA_in_ga(cons(nil, T4), cons(nil, T4)) → gopherA_out_ga(cons(nil, T4), cons(nil, T4))
gopherA_in_ga(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23))) → gopherA_out_ga(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23)))
gopherA_in_ga(cons(cons(cons(T34, T35), T36), T37), T39) → U1_ga(T34, T35, T36, T37, T39, gopherA_in_ga(cons(T34, cons(T35, cons(T36, T37))), T39))
U1_ga(T34, T35, T36, T37, T39, gopherA_out_ga(cons(T34, cons(T35, cons(T36, T37))), T39)) → gopherA_out_ga(cons(cons(cons(T34, T35), T36), T37), T39)

The argument filtering Pi contains the following mapping:
gopherA_in_ga(x1, x2)  =  gopherA_in_ga(x1)
nil  =  nil
gopherA_out_ga(x1, x2)  =  gopherA_out_ga(x2)
cons(x1, x2)  =  cons(x1, x2)
U1_ga(x1, x2, x3, x4, x5, x6)  =  U1_ga(x6)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

Pi-finite rewrite system:
The TRS R consists of the following rules:

gopherA_in_ga(nil, nil) → gopherA_out_ga(nil, nil)
gopherA_in_ga(cons(nil, T4), cons(nil, T4)) → gopherA_out_ga(cons(nil, T4), cons(nil, T4))
gopherA_in_ga(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23))) → gopherA_out_ga(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23)))
gopherA_in_ga(cons(cons(cons(T34, T35), T36), T37), T39) → U1_ga(T34, T35, T36, T37, T39, gopherA_in_ga(cons(T34, cons(T35, cons(T36, T37))), T39))
U1_ga(T34, T35, T36, T37, T39, gopherA_out_ga(cons(T34, cons(T35, cons(T36, T37))), T39)) → gopherA_out_ga(cons(cons(cons(T34, T35), T36), T37), T39)

The argument filtering Pi contains the following mapping:
gopherA_in_ga(x1, x2)  =  gopherA_in_ga(x1)
nil  =  nil
gopherA_out_ga(x1, x2)  =  gopherA_out_ga(x2)
cons(x1, x2)  =  cons(x1, x2)
U1_ga(x1, x2, x3, x4, x5, x6)  =  U1_ga(x6)

(5) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LOPSTR] we result in the following initial DP problem:
Pi DP problem:
The TRS P consists of the following rules:

GOPHERA_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → U1_GA(T34, T35, T36, T37, T39, gopherA_in_ga(cons(T34, cons(T35, cons(T36, T37))), T39))
GOPHERA_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → GOPHERA_IN_GA(cons(T34, cons(T35, cons(T36, T37))), T39)

The TRS R consists of the following rules:

gopherA_in_ga(nil, nil) → gopherA_out_ga(nil, nil)
gopherA_in_ga(cons(nil, T4), cons(nil, T4)) → gopherA_out_ga(cons(nil, T4), cons(nil, T4))
gopherA_in_ga(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23))) → gopherA_out_ga(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23)))
gopherA_in_ga(cons(cons(cons(T34, T35), T36), T37), T39) → U1_ga(T34, T35, T36, T37, T39, gopherA_in_ga(cons(T34, cons(T35, cons(T36, T37))), T39))
U1_ga(T34, T35, T36, T37, T39, gopherA_out_ga(cons(T34, cons(T35, cons(T36, T37))), T39)) → gopherA_out_ga(cons(cons(cons(T34, T35), T36), T37), T39)

The argument filtering Pi contains the following mapping:
gopherA_in_ga(x1, x2)  =  gopherA_in_ga(x1)
nil  =  nil
gopherA_out_ga(x1, x2)  =  gopherA_out_ga(x2)
cons(x1, x2)  =  cons(x1, x2)
U1_ga(x1, x2, x3, x4, x5, x6)  =  U1_ga(x6)
GOPHERA_IN_GA(x1, x2)  =  GOPHERA_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5, x6)  =  U1_GA(x6)

We have to consider all (P,R,Pi)-chains

(6) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

GOPHERA_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → U1_GA(T34, T35, T36, T37, T39, gopherA_in_ga(cons(T34, cons(T35, cons(T36, T37))), T39))
GOPHERA_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → GOPHERA_IN_GA(cons(T34, cons(T35, cons(T36, T37))), T39)

The TRS R consists of the following rules:

gopherA_in_ga(nil, nil) → gopherA_out_ga(nil, nil)
gopherA_in_ga(cons(nil, T4), cons(nil, T4)) → gopherA_out_ga(cons(nil, T4), cons(nil, T4))
gopherA_in_ga(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23))) → gopherA_out_ga(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23)))
gopherA_in_ga(cons(cons(cons(T34, T35), T36), T37), T39) → U1_ga(T34, T35, T36, T37, T39, gopherA_in_ga(cons(T34, cons(T35, cons(T36, T37))), T39))
U1_ga(T34, T35, T36, T37, T39, gopherA_out_ga(cons(T34, cons(T35, cons(T36, T37))), T39)) → gopherA_out_ga(cons(cons(cons(T34, T35), T36), T37), T39)

The argument filtering Pi contains the following mapping:
gopherA_in_ga(x1, x2)  =  gopherA_in_ga(x1)
nil  =  nil
gopherA_out_ga(x1, x2)  =  gopherA_out_ga(x2)
cons(x1, x2)  =  cons(x1, x2)
U1_ga(x1, x2, x3, x4, x5, x6)  =  U1_ga(x6)
GOPHERA_IN_GA(x1, x2)  =  GOPHERA_IN_GA(x1)
U1_GA(x1, x2, x3, x4, x5, x6)  =  U1_GA(x6)

We have to consider all (P,R,Pi)-chains

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 1 less node.

(8) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

GOPHERA_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → GOPHERA_IN_GA(cons(T34, cons(T35, cons(T36, T37))), T39)

The TRS R consists of the following rules:

gopherA_in_ga(nil, nil) → gopherA_out_ga(nil, nil)
gopherA_in_ga(cons(nil, T4), cons(nil, T4)) → gopherA_out_ga(cons(nil, T4), cons(nil, T4))
gopherA_in_ga(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23))) → gopherA_out_ga(cons(cons(nil, T22), T23), cons(nil, cons(T22, T23)))
gopherA_in_ga(cons(cons(cons(T34, T35), T36), T37), T39) → U1_ga(T34, T35, T36, T37, T39, gopherA_in_ga(cons(T34, cons(T35, cons(T36, T37))), T39))
U1_ga(T34, T35, T36, T37, T39, gopherA_out_ga(cons(T34, cons(T35, cons(T36, T37))), T39)) → gopherA_out_ga(cons(cons(cons(T34, T35), T36), T37), T39)

The argument filtering Pi contains the following mapping:
gopherA_in_ga(x1, x2)  =  gopherA_in_ga(x1)
nil  =  nil
gopherA_out_ga(x1, x2)  =  gopherA_out_ga(x2)
cons(x1, x2)  =  cons(x1, x2)
U1_ga(x1, x2, x3, x4, x5, x6)  =  U1_ga(x6)
GOPHERA_IN_GA(x1, x2)  =  GOPHERA_IN_GA(x1)

We have to consider all (P,R,Pi)-chains

(9) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(10) Obligation:

Pi DP problem:
The TRS P consists of the following rules:

GOPHERA_IN_GA(cons(cons(cons(T34, T35), T36), T37), T39) → GOPHERA_IN_GA(cons(T34, cons(T35, cons(T36, T37))), T39)

R is empty.
The argument filtering Pi contains the following mapping:
cons(x1, x2)  =  cons(x1, x2)
GOPHERA_IN_GA(x1, x2)  =  GOPHERA_IN_GA(x1)

We have to consider all (P,R,Pi)-chains

(11) PiDPToQDPProof (SOUND transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

GOPHERA_IN_GA(cons(cons(cons(T34, T35), T36), T37)) → GOPHERA_IN_GA(cons(T34, cons(T35, cons(T36, T37))))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(13) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

GOPHERA_IN_GA(cons(cons(cons(T34, T35), T36), T37)) → GOPHERA_IN_GA(cons(T34, cons(T35, cons(T36, T37))))


Used ordering: Knuth-Bendix order [KBO] with precedence:
cons2 > GOPHERAINGA1

and weight map:

GOPHERA_IN_GA_1=1
cons_2=0

The variable weight is 1

(14) Obligation:

Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(15) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(16) YES