Left Termination of the query pattern p_in_3(a, a, a) w.r.t. the given Prolog program could not be shown:

0 Prolog
↳1 CutEliminatorProof (⇐)
↳2 Prolog
  ↳3 PrologToPiTRSProof (⇐)
    ↳4 PiTRS
    ↳5 DependencyPairsProof (⇔)
    ↳6 PiDP
    ↳7 DependencyGraphProof (⇔)
    ↳8 PiDP
    ↳9 UsableRulesProof (⇔)
    ↳10 PiDP
    ↳11 PiDPToQDPProof (⇐)
    ↳12 QDP
    ↳13 NonTerminationProof (⇔)
    ↳14 FALSE
  ↳15 PrologToPiTRSProof (⇐)
    ↳16 PiTRS
    ↳17 DependencyPairsProof (⇔)
    ↳18 PiDP
    ↳19 DependencyGraphProof (⇔)
    ↳20 PiDP
    ↳21 UsableRulesProof (⇔)
    ↳22 PiDP
    ↳23 PiDPToQDPProof (⇐)
    ↳24 QDP
    ↳25 NonTerminationProof (⇔)
    ↳26 FALSE

(0) Obligation:

Clauses:

p(X, Y, Z) :- ','(append(X, Y, Z), !).
append([], Y, Y).
append(.(H, Xs), Ys, .(H, Zs)) :- append(Xs, Ys, Zs).

Queries:

p(a,a,a).

(1) CutEliminatorProof (SOUND transformation)

Eliminated all cuts by simply ignoring them[PROLOG].

(2) Obligation:

Clauses:

p(X, Y, Z) :- append(X, Y, Z).
append([], Y, Y).
append(.(H, Xs), Ys, .(H, Zs)) :- append(Xs, Ys, Zs).

Queries:

p(a,a,a).

(3) PrologToPiTRSProof (SOUND transformation)

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

p_in_aaa(X, Y, Z) → U1_aaa(X, Y, Z, append_in_aaa(X, Y, Z))
append_in_aaa([], Y, Y) → append_out_aaa([], Y, Y)
append_in_aaa(.(H, Xs), Ys, .(H, Zs)) → U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(H, Xs), Ys, .(H, Zs))
U1_aaa(X, Y, Z, append_out_aaa(X, Y, Z)) → p_out_aaa(X, Y, Z)

The argument filtering Pi contains the following mapping:
p_in_aaa(x1, x2, x3)  =  p_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
.(x1, x2)  =  .(x2)
p_out_aaa(x1, x2, x3)  =  p_out_aaa(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

p_in_aaa(X, Y, Z) → U1_aaa(X, Y, Z, append_in_aaa(X, Y, Z))
append_in_aaa([], Y, Y) → append_out_aaa([], Y, Y)
append_in_aaa(.(H, Xs), Ys, .(H, Zs)) → U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(H, Xs), Ys, .(H, Zs))
U1_aaa(X, Y, Z, append_out_aaa(X, Y, Z)) → p_out_aaa(X, Y, Z)

The argument filtering Pi contains the following mapping:
p_in_aaa(x1, x2, x3)  =  p_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
.(x1, x2)  =  .(x2)
p_out_aaa(x1, x2, x3)  =  p_out_aaa(x1)

(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:

P_IN_AAA(X, Y, Z) → U1_AAA(X, Y, Z, append_in_aaa(X, Y, Z))
P_IN_AAA(X, Y, Z) → APPEND_IN_AAA(X, Y, Z)
APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → U2_AAA(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

p_in_aaa(X, Y, Z) → U1_aaa(X, Y, Z, append_in_aaa(X, Y, Z))
append_in_aaa([], Y, Y) → append_out_aaa([], Y, Y)
append_in_aaa(.(H, Xs), Ys, .(H, Zs)) → U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(H, Xs), Ys, .(H, Zs))
U1_aaa(X, Y, Z, append_out_aaa(X, Y, Z)) → p_out_aaa(X, Y, Z)

The argument filtering Pi contains the following mapping:
p_in_aaa(x1, x2, x3)  =  p_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
.(x1, x2)  =  .(x2)
p_out_aaa(x1, x2, x3)  =  p_out_aaa(x1)
P_IN_AAA(x1, x2, x3)  =  P_IN_AAA
U1_AAA(x1, x2, x3, x4)  =  U1_AAA(x4)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U2_AAA(x1, x2, x3, x4, x5)  =  U2_AAA(x5)

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

(6) Obligation:

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

P_IN_AAA(X, Y, Z) → U1_AAA(X, Y, Z, append_in_aaa(X, Y, Z))
P_IN_AAA(X, Y, Z) → APPEND_IN_AAA(X, Y, Z)
APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → U2_AAA(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

p_in_aaa(X, Y, Z) → U1_aaa(X, Y, Z, append_in_aaa(X, Y, Z))
append_in_aaa([], Y, Y) → append_out_aaa([], Y, Y)
append_in_aaa(.(H, Xs), Ys, .(H, Zs)) → U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(H, Xs), Ys, .(H, Zs))
U1_aaa(X, Y, Z, append_out_aaa(X, Y, Z)) → p_out_aaa(X, Y, Z)

The argument filtering Pi contains the following mapping:
p_in_aaa(x1, x2, x3)  =  p_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
.(x1, x2)  =  .(x2)
p_out_aaa(x1, x2, x3)  =  p_out_aaa(x1)
P_IN_AAA(x1, x2, x3)  =  P_IN_AAA
U1_AAA(x1, x2, x3, x4)  =  U1_AAA(x4)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U2_AAA(x1, x2, x3, x4, x5)  =  U2_AAA(x5)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

p_in_aaa(X, Y, Z) → U1_aaa(X, Y, Z, append_in_aaa(X, Y, Z))
append_in_aaa([], Y, Y) → append_out_aaa([], Y, Y)
append_in_aaa(.(H, Xs), Ys, .(H, Zs)) → U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(H, Xs), Ys, .(H, Zs))
U1_aaa(X, Y, Z, append_out_aaa(X, Y, Z)) → p_out_aaa(X, Y, Z)

The argument filtering Pi contains the following mapping:
p_in_aaa(x1, x2, x3)  =  p_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
.(x1, x2)  =  .(x2)
p_out_aaa(x1, x2, x3)  =  p_out_aaa(x1)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA

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:

APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

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

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:

APPEND_IN_AAA → APPEND_IN_AAA

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

(13) 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 = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APPEND_IN_AAA to APPEND_IN_AAA.



(14) FALSE

(15) PrologToPiTRSProof (SOUND transformation)

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

p_in_aaa(X, Y, Z) → U1_aaa(X, Y, Z, append_in_aaa(X, Y, Z))
append_in_aaa([], Y, Y) → append_out_aaa([], Y, Y)
append_in_aaa(.(H, Xs), Ys, .(H, Zs)) → U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(H, Xs), Ys, .(H, Zs))
U1_aaa(X, Y, Z, append_out_aaa(X, Y, Z)) → p_out_aaa(X, Y, Z)

The argument filtering Pi contains the following mapping:
p_in_aaa(x1, x2, x3)  =  p_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
.(x1, x2)  =  .(x2)
p_out_aaa(x1, x2, x3)  =  p_out_aaa(x1)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) Obligation:

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

p_in_aaa(X, Y, Z) → U1_aaa(X, Y, Z, append_in_aaa(X, Y, Z))
append_in_aaa([], Y, Y) → append_out_aaa([], Y, Y)
append_in_aaa(.(H, Xs), Ys, .(H, Zs)) → U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(H, Xs), Ys, .(H, Zs))
U1_aaa(X, Y, Z, append_out_aaa(X, Y, Z)) → p_out_aaa(X, Y, Z)

The argument filtering Pi contains the following mapping:
p_in_aaa(x1, x2, x3)  =  p_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
.(x1, x2)  =  .(x2)
p_out_aaa(x1, x2, x3)  =  p_out_aaa(x1)

(17) 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:

P_IN_AAA(X, Y, Z) → U1_AAA(X, Y, Z, append_in_aaa(X, Y, Z))
P_IN_AAA(X, Y, Z) → APPEND_IN_AAA(X, Y, Z)
APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → U2_AAA(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

p_in_aaa(X, Y, Z) → U1_aaa(X, Y, Z, append_in_aaa(X, Y, Z))
append_in_aaa([], Y, Y) → append_out_aaa([], Y, Y)
append_in_aaa(.(H, Xs), Ys, .(H, Zs)) → U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(H, Xs), Ys, .(H, Zs))
U1_aaa(X, Y, Z, append_out_aaa(X, Y, Z)) → p_out_aaa(X, Y, Z)

The argument filtering Pi contains the following mapping:
p_in_aaa(x1, x2, x3)  =  p_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
.(x1, x2)  =  .(x2)
p_out_aaa(x1, x2, x3)  =  p_out_aaa(x1)
P_IN_AAA(x1, x2, x3)  =  P_IN_AAA
U1_AAA(x1, x2, x3, x4)  =  U1_AAA(x4)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U2_AAA(x1, x2, x3, x4, x5)  =  U2_AAA(x5)

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

(18) Obligation:

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

P_IN_AAA(X, Y, Z) → U1_AAA(X, Y, Z, append_in_aaa(X, Y, Z))
P_IN_AAA(X, Y, Z) → APPEND_IN_AAA(X, Y, Z)
APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → U2_AAA(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

p_in_aaa(X, Y, Z) → U1_aaa(X, Y, Z, append_in_aaa(X, Y, Z))
append_in_aaa([], Y, Y) → append_out_aaa([], Y, Y)
append_in_aaa(.(H, Xs), Ys, .(H, Zs)) → U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(H, Xs), Ys, .(H, Zs))
U1_aaa(X, Y, Z, append_out_aaa(X, Y, Z)) → p_out_aaa(X, Y, Z)

The argument filtering Pi contains the following mapping:
p_in_aaa(x1, x2, x3)  =  p_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
.(x1, x2)  =  .(x2)
p_out_aaa(x1, x2, x3)  =  p_out_aaa(x1)
P_IN_AAA(x1, x2, x3)  =  P_IN_AAA
U1_AAA(x1, x2, x3, x4)  =  U1_AAA(x4)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA
U2_AAA(x1, x2, x3, x4, x5)  =  U2_AAA(x5)

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

(19) DependencyGraphProof (EQUIVALENT transformation)

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

(20) Obligation:

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

APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

The TRS R consists of the following rules:

p_in_aaa(X, Y, Z) → U1_aaa(X, Y, Z, append_in_aaa(X, Y, Z))
append_in_aaa([], Y, Y) → append_out_aaa([], Y, Y)
append_in_aaa(.(H, Xs), Ys, .(H, Zs)) → U2_aaa(H, Xs, Ys, Zs, append_in_aaa(Xs, Ys, Zs))
U2_aaa(H, Xs, Ys, Zs, append_out_aaa(Xs, Ys, Zs)) → append_out_aaa(.(H, Xs), Ys, .(H, Zs))
U1_aaa(X, Y, Z, append_out_aaa(X, Y, Z)) → p_out_aaa(X, Y, Z)

The argument filtering Pi contains the following mapping:
p_in_aaa(x1, x2, x3)  =  p_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append_in_aaa(x1, x2, x3)  =  append_in_aaa
append_out_aaa(x1, x2, x3)  =  append_out_aaa(x1)
U2_aaa(x1, x2, x3, x4, x5)  =  U2_aaa(x5)
.(x1, x2)  =  .(x2)
p_out_aaa(x1, x2, x3)  =  p_out_aaa(x1)
APPEND_IN_AAA(x1, x2, x3)  =  APPEND_IN_AAA

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

(21) UsableRulesProof (EQUIVALENT transformation)

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

(22) Obligation:

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

APPEND_IN_AAA(.(H, Xs), Ys, .(H, Zs)) → APPEND_IN_AAA(Xs, Ys, Zs)

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

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

(23) PiDPToQDPProof (SOUND transformation)

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

(24) Obligation:

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

APPEND_IN_AAA → APPEND_IN_AAA

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

(25) 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 = APPEND_IN_AAA evaluates to t =APPEND_IN_AAA

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Semiunifier: [ ]
  • Matcher: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APPEND_IN_AAA to APPEND_IN_AAA.



(26) FALSE