Left Termination of the query pattern app_in_3(g, a, a) w.r.t. the given Prolog program could successfully be proven:

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 QDPSizeChangeProof (⇔)
↳14 YES

(0) Obligation:

Clauses:

app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).

Queries:

app(g,a,a).

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph.

(2) Obligation:

Clauses:

app1([], T5, T5).
app1(.(T10, []), T20, .(T10, T20)).
app1(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) :- app1(T30, T33, T34).

Queries:

app1(g,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:
app1_in: (b,f,f)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

app1_in_gaa([], T5, T5) → app1_out_gaa([], T5, T5)
app1_in_gaa(.(T10, []), T20, .(T10, T20)) → app1_out_gaa(.(T10, []), T20, .(T10, T20))
app1_in_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_gaa(T10, T29, T30, T33, T34, app1_in_gaa(T30, T33, T34))
U1_gaa(T10, T29, T30, T33, T34, app1_out_gaa(T30, T33, T34)) → app1_out_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34)))

The argument filtering Pi contains the following mapping:
app1_in_gaa(x1, x2, x3)  =  app1_in_gaa(x1)
[]  =  []
app1_out_gaa(x1, x2, x3)  =  app1_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(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:

app1_in_gaa([], T5, T5) → app1_out_gaa([], T5, T5)
app1_in_gaa(.(T10, []), T20, .(T10, T20)) → app1_out_gaa(.(T10, []), T20, .(T10, T20))
app1_in_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_gaa(T10, T29, T30, T33, T34, app1_in_gaa(T30, T33, T34))
U1_gaa(T10, T29, T30, T33, T34, app1_out_gaa(T30, T33, T34)) → app1_out_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34)))

The argument filtering Pi contains the following mapping:
app1_in_gaa(x1, x2, x3)  =  app1_in_gaa(x1)
[]  =  []
app1_out_gaa(x1, x2, x3)  =  app1_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(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:

APP1_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_GAA(T10, T29, T30, T33, T34, app1_in_gaa(T30, T33, T34))
APP1_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → APP1_IN_GAA(T30, T33, T34)

The TRS R consists of the following rules:

app1_in_gaa([], T5, T5) → app1_out_gaa([], T5, T5)
app1_in_gaa(.(T10, []), T20, .(T10, T20)) → app1_out_gaa(.(T10, []), T20, .(T10, T20))
app1_in_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_gaa(T10, T29, T30, T33, T34, app1_in_gaa(T30, T33, T34))
U1_gaa(T10, T29, T30, T33, T34, app1_out_gaa(T30, T33, T34)) → app1_out_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34)))

The argument filtering Pi contains the following mapping:
app1_in_gaa(x1, x2, x3)  =  app1_in_gaa(x1)
[]  =  []
app1_out_gaa(x1, x2, x3)  =  app1_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
APP1_IN_GAA(x1, x2, x3)  =  APP1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(x6)

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

(6) Obligation:

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

APP1_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_GAA(T10, T29, T30, T33, T34, app1_in_gaa(T30, T33, T34))
APP1_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → APP1_IN_GAA(T30, T33, T34)

The TRS R consists of the following rules:

app1_in_gaa([], T5, T5) → app1_out_gaa([], T5, T5)
app1_in_gaa(.(T10, []), T20, .(T10, T20)) → app1_out_gaa(.(T10, []), T20, .(T10, T20))
app1_in_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_gaa(T10, T29, T30, T33, T34, app1_in_gaa(T30, T33, T34))
U1_gaa(T10, T29, T30, T33, T34, app1_out_gaa(T30, T33, T34)) → app1_out_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34)))

The argument filtering Pi contains the following mapping:
app1_in_gaa(x1, x2, x3)  =  app1_in_gaa(x1)
[]  =  []
app1_out_gaa(x1, x2, x3)  =  app1_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
APP1_IN_GAA(x1, x2, x3)  =  APP1_IN_GAA(x1)
U1_GAA(x1, x2, x3, x4, x5, x6)  =  U1_GAA(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:

APP1_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → APP1_IN_GAA(T30, T33, T34)

The TRS R consists of the following rules:

app1_in_gaa([], T5, T5) → app1_out_gaa([], T5, T5)
app1_in_gaa(.(T10, []), T20, .(T10, T20)) → app1_out_gaa(.(T10, []), T20, .(T10, T20))
app1_in_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → U1_gaa(T10, T29, T30, T33, T34, app1_in_gaa(T30, T33, T34))
U1_gaa(T10, T29, T30, T33, T34, app1_out_gaa(T30, T33, T34)) → app1_out_gaa(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34)))

The argument filtering Pi contains the following mapping:
app1_in_gaa(x1, x2, x3)  =  app1_in_gaa(x1)
[]  =  []
app1_out_gaa(x1, x2, x3)  =  app1_out_gaa
.(x1, x2)  =  .(x1, x2)
U1_gaa(x1, x2, x3, x4, x5, x6)  =  U1_gaa(x6)
APP1_IN_GAA(x1, x2, x3)  =  APP1_IN_GAA(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:

APP1_IN_GAA(.(T10, .(T29, T30)), T33, .(T10, .(T29, T34))) → APP1_IN_GAA(T30, T33, T34)

R is empty.
The argument filtering Pi contains the following mapping:
.(x1, x2)  =  .(x1, x2)
APP1_IN_GAA(x1, x2, x3)  =  APP1_IN_GAA(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:

APP1_IN_GAA(.(T10, .(T29, T30))) → APP1_IN_GAA(T30)

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

(13) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • APP1_IN_GAA(.(T10, .(T29, T30))) → APP1_IN_GAA(T30)
    The graph contains the following edges 1 > 1

(14) YES