Left Termination of the query pattern reverse_in_2(a, g) w.r.t. the given Prolog program could not be shown:

0 Prolog
  ↳1 PrologToPiTRSProof (⇐)
    ↳2 PiTRS
    ↳3 DependencyPairsProof (⇔)
    ↳4 PiDP
    ↳5 DependencyGraphProof (⇔)
    ↳6 PiDP
    ↳7 UsableRulesProof (⇔)
    ↳8 PiDP
    ↳9 PiDPToQDPProof (⇐)
    ↳10 QDP
      ↳11 Instantiation (⇔)
        ↳12 QDP
      ↳13 Instantiation (⇔)
        ↳14 QDP
        ↳15 Instantiation (⇔)
        ↳16 QDP
        ↳17 NonTerminationProof (⇔)
        ↳18 FALSE
  ↳19 PrologToPiTRSProof (⇐)
    ↳20 PiTRS
    ↳21 DependencyPairsProof (⇔)
    ↳22 PiDP
    ↳23 DependencyGraphProof (⇔)
    ↳24 PiDP
    ↳25 UsableRulesProof (⇔)
    ↳26 PiDP
    ↳27 PiDPToQDPProof (⇐)
    ↳28 QDP
      ↳29 Instantiation (⇔)
        ↳30 QDP
      ↳31 Instantiation (⇔)
        ↳32 QDP
        ↳33 Instantiation (⇔)
        ↳34 QDP
        ↳35 NonTerminationProof (⇔)
        ↳36 FALSE

(0) Obligation:

Clauses:

reverse(X1s, X2s) :- reverse(X1s, [], X2s).
reverse([], Xs, Xs).
reverse(.(X, X1s), X2s, Ys) :- reverse(X1s, .(X, X2s), Ys).

Queries:

reverse(a,g).

(1) PrologToPiTRSProof (SOUND transformation)

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

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(2) Obligation:

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

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)

(3) 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:

REVERSE_IN_AG(X1s, X2s) → U1_AG(X1s, X2s, reverse_in_agg(X1s, [], X2s))
REVERSE_IN_AG(X1s, X2s) → REVERSE_IN_AGG(X1s, [], X2s)
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → U2_AGG(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x3)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x5)

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

(4) Obligation:

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

REVERSE_IN_AG(X1s, X2s) → U1_AG(X1s, X2s, reverse_in_agg(X1s, [], X2s))
REVERSE_IN_AG(X1s, X2s) → REVERSE_IN_AGG(X1s, [], X2s)
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → U2_AGG(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x3)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x5)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

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

(7) UsableRulesProof (EQUIVALENT transformation)

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

(8) Obligation:

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

REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

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

(9) PiDPToQDPProof (SOUND transformation)

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

(10) Obligation:

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

REVERSE_IN_AGG(X2s, Ys) → REVERSE_IN_AGG(.(X2s), Ys)

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

(11) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule REVERSE_IN_AGG(X2s, Ys) → REVERSE_IN_AGG(.(X2s), Ys) we obtained the following new rules [LPAR04]:

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)

(12) Obligation:

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

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)

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

(13) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule REVERSE_IN_AGG(X2s, Ys) → REVERSE_IN_AGG(.(X2s), Ys) we obtained the following new rules [LPAR04]:

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)

(14) Obligation:

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

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)

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

(15) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1) we obtained the following new rules [LPAR04]:

REVERSE_IN_AGG(.(.(z0)), z1) → REVERSE_IN_AGG(.(.(.(z0))), z1)

(16) Obligation:

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

REVERSE_IN_AGG(.(.(z0)), z1) → REVERSE_IN_AGG(.(.(.(z0))), z1)

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

(17) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = REVERSE_IN_AGG(.(.(z0)), z1) evaluates to t =REVERSE_IN_AGG(.(.(.(z0))), z1)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [z0 / .(z0)]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from REVERSE_IN_AGG(.(.(z0)), z1) to REVERSE_IN_AGG(.(.(.(z0))), z1).



(18) FALSE

(19) PrologToPiTRSProof (SOUND transformation)

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

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x3, x4, x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(20) Obligation:

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

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x3, x4, x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)

(21) 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:

REVERSE_IN_AG(X1s, X2s) → U1_AG(X1s, X2s, reverse_in_agg(X1s, [], X2s))
REVERSE_IN_AG(X1s, X2s) → REVERSE_IN_AGG(X1s, [], X2s)
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → U2_AGG(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x3, x4, x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x3, x4, x5)

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

(22) Obligation:

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

REVERSE_IN_AG(X1s, X2s) → U1_AG(X1s, X2s, reverse_in_agg(X1s, [], X2s))
REVERSE_IN_AG(X1s, X2s) → REVERSE_IN_AGG(X1s, [], X2s)
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → U2_AGG(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x3, x4, x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
REVERSE_IN_AG(x1, x2)  =  REVERSE_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)
U2_AGG(x1, x2, x3, x4, x5)  =  U2_AGG(x3, x4, x5)

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

(23) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 1 SCC with 3 less nodes.

(24) Obligation:

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

REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

The TRS R consists of the following rules:

reverse_in_ag(X1s, X2s) → U1_ag(X1s, X2s, reverse_in_agg(X1s, [], X2s))
reverse_in_agg([], Xs, Xs) → reverse_out_agg([], Xs, Xs)
reverse_in_agg(.(X, X1s), X2s, Ys) → U2_agg(X, X1s, X2s, Ys, reverse_in_agg(X1s, .(X, X2s), Ys))
U2_agg(X, X1s, X2s, Ys, reverse_out_agg(X1s, .(X, X2s), Ys)) → reverse_out_agg(.(X, X1s), X2s, Ys)
U1_ag(X1s, X2s, reverse_out_agg(X1s, [], X2s)) → reverse_out_ag(X1s, X2s)

The argument filtering Pi contains the following mapping:
reverse_in_ag(x1, x2)  =  reverse_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
reverse_in_agg(x1, x2, x3)  =  reverse_in_agg(x2, x3)
reverse_out_agg(x1, x2, x3)  =  reverse_out_agg(x1, x2, x3)
U2_agg(x1, x2, x3, x4, x5)  =  U2_agg(x3, x4, x5)
.(x1, x2)  =  .(x2)
[]  =  []
reverse_out_ag(x1, x2)  =  reverse_out_ag(x1, x2)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

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

(25) UsableRulesProof (EQUIVALENT transformation)

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

(26) Obligation:

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

REVERSE_IN_AGG(.(X, X1s), X2s, Ys) → REVERSE_IN_AGG(X1s, .(X, X2s), Ys)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x2)
REVERSE_IN_AGG(x1, x2, x3)  =  REVERSE_IN_AGG(x2, x3)

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

(27) PiDPToQDPProof (SOUND transformation)

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

(28) Obligation:

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

REVERSE_IN_AGG(X2s, Ys) → REVERSE_IN_AGG(.(X2s), Ys)

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

(29) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule REVERSE_IN_AGG(X2s, Ys) → REVERSE_IN_AGG(.(X2s), Ys) we obtained the following new rules [LPAR04]:

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)

(30) Obligation:

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

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)

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

(31) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule REVERSE_IN_AGG(X2s, Ys) → REVERSE_IN_AGG(.(X2s), Ys) we obtained the following new rules [LPAR04]:

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)

(32) Obligation:

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

REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1)

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

(33) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule REVERSE_IN_AGG(.(z0), z1) → REVERSE_IN_AGG(.(.(z0)), z1) we obtained the following new rules [LPAR04]:

REVERSE_IN_AGG(.(.(z0)), z1) → REVERSE_IN_AGG(.(.(.(z0))), z1)

(34) Obligation:

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

REVERSE_IN_AGG(.(.(z0)), z1) → REVERSE_IN_AGG(.(.(.(z0))), z1)

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

(35) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = REVERSE_IN_AGG(.(.(z0)), z1) evaluates to t =REVERSE_IN_AGG(.(.(.(z0))), z1)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [z0 / .(z0)]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from REVERSE_IN_AGG(.(.(z0)), z1) to REVERSE_IN_AGG(.(.(.(z0))), z1).



(36) FALSE