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 PrologToPrologProblemTransformerProof (⇐)
↳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 NO
    ↳15 PrologToPiTRSProof (⇐)
        ↳16 PiTRS
        ↳17 DependencyPairsProof (⇔)
        ↳18 PiDP
        ↳19 DependencyGraphProof (⇔)
        ↳20 PiDP
        ↳21 UsableRulesProof (⇔)
        ↳22 PiDP
        ↳23 PiDPToQDPProof (⇐)
        ↳24 QDP
        ↳25 NonTerminationProof (⇔)
        ↳26 NO

(0) Obligation:

Clauses:

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

Queries:

p(a,a,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

append3(T27, .(T27, T26), T25) :- append3(T27, T26, T25).
p1(T12, T11, T10) :- append3(T12, T11, T10).

Queries:

p1(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:
p1_in: (f,f,f)
append3_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_aaa(T12, T11, T10) → U2_aaa(T12, T11, T10, append3_in_aaa(T12, T11, T10))
append3_in_aaa(T27, .(T27, T26), T25) → U1_aaa(T27, T26, T25, append3_in_aaa(T27, T26, T25))
U1_aaa(T27, T26, T25, append3_out_aaa(T27, T26, T25)) → append3_out_aaa(T27, .(T27, T26), T25)
U2_aaa(T12, T11, T10, append3_out_aaa(T12, T11, T10)) → p1_out_aaa(T12, T11, T10)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
U2_aaa(x1, x2, x3, x4)  =  U2_aaa(x4)
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append3_out_aaa(x1, x2, x3)  =  append3_out_aaa(x1, x2, x3)
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

p1_in_aaa(T12, T11, T10) → U2_aaa(T12, T11, T10, append3_in_aaa(T12, T11, T10))
append3_in_aaa(T27, .(T27, T26), T25) → U1_aaa(T27, T26, T25, append3_in_aaa(T27, T26, T25))
U1_aaa(T27, T26, T25, append3_out_aaa(T27, T26, T25)) → append3_out_aaa(T27, .(T27, T26), T25)
U2_aaa(T12, T11, T10, append3_out_aaa(T12, T11, T10)) → p1_out_aaa(T12, T11, T10)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
U2_aaa(x1, x2, x3, x4)  =  U2_aaa(x4)
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append3_out_aaa(x1, x2, x3)  =  append3_out_aaa(x1, x2, x3)
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)

(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:

P1_IN_AAA(T12, T11, T10) → U2_AAA(T12, T11, T10, append3_in_aaa(T12, T11, T10))
P1_IN_AAA(T12, T11, T10) → APPEND3_IN_AAA(T12, T11, T10)
APPEND3_IN_AAA(T27, .(T27, T26), T25) → U1_AAA(T27, T26, T25, append3_in_aaa(T27, T26, T25))
APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

The TRS R consists of the following rules:

p1_in_aaa(T12, T11, T10) → U2_aaa(T12, T11, T10, append3_in_aaa(T12, T11, T10))
append3_in_aaa(T27, .(T27, T26), T25) → U1_aaa(T27, T26, T25, append3_in_aaa(T27, T26, T25))
U1_aaa(T27, T26, T25, append3_out_aaa(T27, T26, T25)) → append3_out_aaa(T27, .(T27, T26), T25)
U2_aaa(T12, T11, T10, append3_out_aaa(T12, T11, T10)) → p1_out_aaa(T12, T11, T10)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
U2_aaa(x1, x2, x3, x4)  =  U2_aaa(x4)
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append3_out_aaa(x1, x2, x3)  =  append3_out_aaa(x1, x2, x3)
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
P1_IN_AAA(x1, x2, x3)  =  P1_IN_AAA
U2_AAA(x1, x2, x3, x4)  =  U2_AAA(x4)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_IN_AAA
U1_AAA(x1, x2, x3, x4)  =  U1_AAA(x4)

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

(6) Obligation:

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

P1_IN_AAA(T12, T11, T10) → U2_AAA(T12, T11, T10, append3_in_aaa(T12, T11, T10))
P1_IN_AAA(T12, T11, T10) → APPEND3_IN_AAA(T12, T11, T10)
APPEND3_IN_AAA(T27, .(T27, T26), T25) → U1_AAA(T27, T26, T25, append3_in_aaa(T27, T26, T25))
APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

The TRS R consists of the following rules:

p1_in_aaa(T12, T11, T10) → U2_aaa(T12, T11, T10, append3_in_aaa(T12, T11, T10))
append3_in_aaa(T27, .(T27, T26), T25) → U1_aaa(T27, T26, T25, append3_in_aaa(T27, T26, T25))
U1_aaa(T27, T26, T25, append3_out_aaa(T27, T26, T25)) → append3_out_aaa(T27, .(T27, T26), T25)
U2_aaa(T12, T11, T10, append3_out_aaa(T12, T11, T10)) → p1_out_aaa(T12, T11, T10)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
U2_aaa(x1, x2, x3, x4)  =  U2_aaa(x4)
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append3_out_aaa(x1, x2, x3)  =  append3_out_aaa(x1, x2, x3)
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
P1_IN_AAA(x1, x2, x3)  =  P1_IN_AAA
U2_AAA(x1, x2, x3, x4)  =  U2_AAA(x4)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_IN_AAA
U1_AAA(x1, x2, x3, x4)  =  U1_AAA(x4)

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:

APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

The TRS R consists of the following rules:

p1_in_aaa(T12, T11, T10) → U2_aaa(T12, T11, T10, append3_in_aaa(T12, T11, T10))
append3_in_aaa(T27, .(T27, T26), T25) → U1_aaa(T27, T26, T25, append3_in_aaa(T27, T26, T25))
U1_aaa(T27, T26, T25, append3_out_aaa(T27, T26, T25)) → append3_out_aaa(T27, .(T27, T26), T25)
U2_aaa(T12, T11, T10, append3_out_aaa(T12, T11, T10)) → p1_out_aaa(T12, T11, T10)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
U2_aaa(x1, x2, x3, x4)  =  U2_aaa(x4)
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append3_out_aaa(x1, x2, x3)  =  append3_out_aaa(x1, x2, x3)
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_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:

APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_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:

APPEND3_IN_AAA → APPEND3_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 = APPEND3_IN_AAA evaluates to t =APPEND3_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 APPEND3_IN_AAA to APPEND3_IN_AAA.



(14) NO

(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:
p1_in: (f,f,f)
append3_in: (f,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

p1_in_aaa(T12, T11, T10) → U2_aaa(T12, T11, T10, append3_in_aaa(T12, T11, T10))
append3_in_aaa(T27, .(T27, T26), T25) → U1_aaa(T27, T26, T25, append3_in_aaa(T27, T26, T25))
U1_aaa(T27, T26, T25, append3_out_aaa(T27, T26, T25)) → append3_out_aaa(T27, .(T27, T26), T25)
U2_aaa(T12, T11, T10, append3_out_aaa(T12, T11, T10)) → p1_out_aaa(T12, T11, T10)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
U2_aaa(x1, x2, x3, x4)  =  U2_aaa(x4)
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append3_out_aaa(x1, x2, x3)  =  append3_out_aaa(x1, x2, x3)
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(16) Obligation:

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

p1_in_aaa(T12, T11, T10) → U2_aaa(T12, T11, T10, append3_in_aaa(T12, T11, T10))
append3_in_aaa(T27, .(T27, T26), T25) → U1_aaa(T27, T26, T25, append3_in_aaa(T27, T26, T25))
U1_aaa(T27, T26, T25, append3_out_aaa(T27, T26, T25)) → append3_out_aaa(T27, .(T27, T26), T25)
U2_aaa(T12, T11, T10, append3_out_aaa(T12, T11, T10)) → p1_out_aaa(T12, T11, T10)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
U2_aaa(x1, x2, x3, x4)  =  U2_aaa(x4)
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append3_out_aaa(x1, x2, x3)  =  append3_out_aaa(x1, x2, x3)
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)

(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:

P1_IN_AAA(T12, T11, T10) → U2_AAA(T12, T11, T10, append3_in_aaa(T12, T11, T10))
P1_IN_AAA(T12, T11, T10) → APPEND3_IN_AAA(T12, T11, T10)
APPEND3_IN_AAA(T27, .(T27, T26), T25) → U1_AAA(T27, T26, T25, append3_in_aaa(T27, T26, T25))
APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

The TRS R consists of the following rules:

p1_in_aaa(T12, T11, T10) → U2_aaa(T12, T11, T10, append3_in_aaa(T12, T11, T10))
append3_in_aaa(T27, .(T27, T26), T25) → U1_aaa(T27, T26, T25, append3_in_aaa(T27, T26, T25))
U1_aaa(T27, T26, T25, append3_out_aaa(T27, T26, T25)) → append3_out_aaa(T27, .(T27, T26), T25)
U2_aaa(T12, T11, T10, append3_out_aaa(T12, T11, T10)) → p1_out_aaa(T12, T11, T10)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
U2_aaa(x1, x2, x3, x4)  =  U2_aaa(x4)
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append3_out_aaa(x1, x2, x3)  =  append3_out_aaa(x1, x2, x3)
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
P1_IN_AAA(x1, x2, x3)  =  P1_IN_AAA
U2_AAA(x1, x2, x3, x4)  =  U2_AAA(x4)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_IN_AAA
U1_AAA(x1, x2, x3, x4)  =  U1_AAA(x4)

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

(18) Obligation:

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

P1_IN_AAA(T12, T11, T10) → U2_AAA(T12, T11, T10, append3_in_aaa(T12, T11, T10))
P1_IN_AAA(T12, T11, T10) → APPEND3_IN_AAA(T12, T11, T10)
APPEND3_IN_AAA(T27, .(T27, T26), T25) → U1_AAA(T27, T26, T25, append3_in_aaa(T27, T26, T25))
APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

The TRS R consists of the following rules:

p1_in_aaa(T12, T11, T10) → U2_aaa(T12, T11, T10, append3_in_aaa(T12, T11, T10))
append3_in_aaa(T27, .(T27, T26), T25) → U1_aaa(T27, T26, T25, append3_in_aaa(T27, T26, T25))
U1_aaa(T27, T26, T25, append3_out_aaa(T27, T26, T25)) → append3_out_aaa(T27, .(T27, T26), T25)
U2_aaa(T12, T11, T10, append3_out_aaa(T12, T11, T10)) → p1_out_aaa(T12, T11, T10)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
U2_aaa(x1, x2, x3, x4)  =  U2_aaa(x4)
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append3_out_aaa(x1, x2, x3)  =  append3_out_aaa(x1, x2, x3)
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
P1_IN_AAA(x1, x2, x3)  =  P1_IN_AAA
U2_AAA(x1, x2, x3, x4)  =  U2_AAA(x4)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_IN_AAA
U1_AAA(x1, x2, x3, x4)  =  U1_AAA(x4)

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:

APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

The TRS R consists of the following rules:

p1_in_aaa(T12, T11, T10) → U2_aaa(T12, T11, T10, append3_in_aaa(T12, T11, T10))
append3_in_aaa(T27, .(T27, T26), T25) → U1_aaa(T27, T26, T25, append3_in_aaa(T27, T26, T25))
U1_aaa(T27, T26, T25, append3_out_aaa(T27, T26, T25)) → append3_out_aaa(T27, .(T27, T26), T25)
U2_aaa(T12, T11, T10, append3_out_aaa(T12, T11, T10)) → p1_out_aaa(T12, T11, T10)

The argument filtering Pi contains the following mapping:
p1_in_aaa(x1, x2, x3)  =  p1_in_aaa
U2_aaa(x1, x2, x3, x4)  =  U2_aaa(x4)
append3_in_aaa(x1, x2, x3)  =  append3_in_aaa
U1_aaa(x1, x2, x3, x4)  =  U1_aaa(x4)
append3_out_aaa(x1, x2, x3)  =  append3_out_aaa(x1, x2, x3)
p1_out_aaa(x1, x2, x3)  =  p1_out_aaa(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_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:

APPEND3_IN_AAA(T27, .(T27, T26), T25) → APPEND3_IN_AAA(T27, T26, T25)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APPEND3_IN_AAA(x1, x2, x3)  =  APPEND3_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:

APPEND3_IN_AAA → APPEND3_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 = APPEND3_IN_AAA evaluates to t =APPEND3_IN_AAA

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



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from APPEND3_IN_AAA to APPEND3_IN_AAA.



(26) NO