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

(0) Obligation:

Clauses:

prefix(Xs, Ys) :- app(Xs, X1, Ys).
app([], X, X).
app(.(X, Xs), Ys, .(X, Zs)) :- app(Xs, Ys, Zs).

Query: prefix(g,a)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
3 [label="A: prefix(T1, T2)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
5 [label="5: prefix(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="B: app(T5, X6, T7)\n\nT5 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: app(T5, X6, T7), SCOPE: 2, CLAUSE: 1 | app(T5, X6, T7), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: app(T5, X6, T7), SCOPE: 2, CLAUSE: 1\n\nT5 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: app(T5, X6, T7), SCOPE: 2, CLAUSE: 2\n\nT5 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: app(T20, X31, T22)\n\nT20 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 26 [arrowhead = none ];
26 -> 5;

27 [label="ONLY EVAL with clause\nprefix(X4, X5) :- app(X4, X6, X5).\nand substitutionT1 -> T5,\nX4 -> T5,\nT2 -> T7,\nX5 -> T7,\nT6 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 27 [arrowhead = none ];
27 -> 8;

28 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 28 [arrowhead = none ];
28 -> 12;

29 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
12 -> 29 [arrowhead = none ];
29 -> 16;

30 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
12 -> 30 [arrowhead = none ];
30 -> 17;

31 [label="EVAL with clause\napp([], X15, X15).\nand substitutionT5 -> [],\nX6 -> T12,\nX15 -> T12,\nT7 -> T12,\nX16 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
16 -> 31 [arrowhead = none ];
31 -> 19;

32 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 32 [arrowhead = none ];
32 -> 20;

33 [label="EVAL with clause\napp(.(X27, X28), X29, .(X27, X30)) :- app(X28, X29, X30).\nand substitutionX27 -> T19,\nX28 -> T20,\nT5 -> .(T19, T20),\nX6 -> X31,\nX29 -> X31,\nX30 -> T22,\nT7 -> .(T19, T22),\nT21 -> T22", fontsize=14, style = filled, fillcolor = lightblue];
17 -> 33 [arrowhead = none ];
33 -> 24;

34 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 34 [arrowhead = none ];
34 -> 25;

35 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
19 -> 35 [arrowhead = none ];
35 -> 21;

36 [label="INSTANCE with matching:\nT5 -> T20\nX6 -> X31\nT7 -> T22", fontsize=14, style = filled, fillcolor = lightgrey];
24 -> 36 [arrowhead = none , style=dashed];
36 -> 8[style=dashed];

}

(2) Obligation:

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

prefixA_in_ga(T5, T7) → U1_ga(T5, T7, appB_in_gaa(T5, X6, T7))
appB_in_gaa([], T12, T12) → appB_out_gaa([], T12, T12)
appB_in_gaa(.(T19, T20), X31, .(T19, T22)) → U2_gaa(T19, T20, X31, T22, appB_in_gaa(T20, X31, T22))
U2_gaa(T19, T20, X31, T22, appB_out_gaa(T20, X31, T22)) → appB_out_gaa(.(T19, T20), X31, .(T19, T22))
U1_ga(T5, T7, appB_out_gaa(T5, X6, T7)) → prefixA_out_ga(T5, T7)

The argument filtering Pi contains the following mapping:
prefixA_in_ga(x1, x2)  =  prefixA_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
appB_in_gaa(x1, x2, x3)  =  appB_in_gaa(x1)
[]  =  []
appB_out_gaa(x1, x2, x3)  =  appB_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x2, x5)
prefixA_out_ga(x1, x2)  =  prefixA_out_ga(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:

PREFIXA_IN_GA(T5, T7) → U1_GA(T5, T7, appB_in_gaa(T5, X6, T7))
PREFIXA_IN_GA(T5, T7) → APPB_IN_GAA(T5, X6, T7)
APPB_IN_GAA(.(T19, T20), X31, .(T19, T22)) → U2_GAA(T19, T20, X31, T22, appB_in_gaa(T20, X31, T22))
APPB_IN_GAA(.(T19, T20), X31, .(T19, T22)) → APPB_IN_GAA(T20, X31, T22)

The TRS R consists of the following rules:

prefixA_in_ga(T5, T7) → U1_ga(T5, T7, appB_in_gaa(T5, X6, T7))
appB_in_gaa([], T12, T12) → appB_out_gaa([], T12, T12)
appB_in_gaa(.(T19, T20), X31, .(T19, T22)) → U2_gaa(T19, T20, X31, T22, appB_in_gaa(T20, X31, T22))
U2_gaa(T19, T20, X31, T22, appB_out_gaa(T20, X31, T22)) → appB_out_gaa(.(T19, T20), X31, .(T19, T22))
U1_ga(T5, T7, appB_out_gaa(T5, X6, T7)) → prefixA_out_ga(T5, T7)

The argument filtering Pi contains the following mapping:
prefixA_in_ga(x1, x2)  =  prefixA_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
appB_in_gaa(x1, x2, x3)  =  appB_in_gaa(x1)
[]  =  []
appB_out_gaa(x1, x2, x3)  =  appB_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x2, x5)
prefixA_out_ga(x1, x2)  =  prefixA_out_ga(x1)
PREFIXA_IN_GA(x1, x2)  =  PREFIXA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APPB_IN_GAA(x1, x2, x3)  =  APPB_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4, x5)  =  U2_GAA(x1, x2, x5)

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

(4) Obligation:

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

PREFIXA_IN_GA(T5, T7) → U1_GA(T5, T7, appB_in_gaa(T5, X6, T7))
PREFIXA_IN_GA(T5, T7) → APPB_IN_GAA(T5, X6, T7)
APPB_IN_GAA(.(T19, T20), X31, .(T19, T22)) → U2_GAA(T19, T20, X31, T22, appB_in_gaa(T20, X31, T22))
APPB_IN_GAA(.(T19, T20), X31, .(T19, T22)) → APPB_IN_GAA(T20, X31, T22)

The TRS R consists of the following rules:

prefixA_in_ga(T5, T7) → U1_ga(T5, T7, appB_in_gaa(T5, X6, T7))
appB_in_gaa([], T12, T12) → appB_out_gaa([], T12, T12)
appB_in_gaa(.(T19, T20), X31, .(T19, T22)) → U2_gaa(T19, T20, X31, T22, appB_in_gaa(T20, X31, T22))
U2_gaa(T19, T20, X31, T22, appB_out_gaa(T20, X31, T22)) → appB_out_gaa(.(T19, T20), X31, .(T19, T22))
U1_ga(T5, T7, appB_out_gaa(T5, X6, T7)) → prefixA_out_ga(T5, T7)

The argument filtering Pi contains the following mapping:
prefixA_in_ga(x1, x2)  =  prefixA_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
appB_in_gaa(x1, x2, x3)  =  appB_in_gaa(x1)
[]  =  []
appB_out_gaa(x1, x2, x3)  =  appB_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x2, x5)
prefixA_out_ga(x1, x2)  =  prefixA_out_ga(x1)
PREFIXA_IN_GA(x1, x2)  =  PREFIXA_IN_GA(x1)
U1_GA(x1, x2, x3)  =  U1_GA(x1, x3)
APPB_IN_GAA(x1, x2, x3)  =  APPB_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4, x5)  =  U2_GAA(x1, x2, 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:

APPB_IN_GAA(.(T19, T20), X31, .(T19, T22)) → APPB_IN_GAA(T20, X31, T22)

The TRS R consists of the following rules:

prefixA_in_ga(T5, T7) → U1_ga(T5, T7, appB_in_gaa(T5, X6, T7))
appB_in_gaa([], T12, T12) → appB_out_gaa([], T12, T12)
appB_in_gaa(.(T19, T20), X31, .(T19, T22)) → U2_gaa(T19, T20, X31, T22, appB_in_gaa(T20, X31, T22))
U2_gaa(T19, T20, X31, T22, appB_out_gaa(T20, X31, T22)) → appB_out_gaa(.(T19, T20), X31, .(T19, T22))
U1_ga(T5, T7, appB_out_gaa(T5, X6, T7)) → prefixA_out_ga(T5, T7)

The argument filtering Pi contains the following mapping:
prefixA_in_ga(x1, x2)  =  prefixA_in_ga(x1)
U1_ga(x1, x2, x3)  =  U1_ga(x1, x3)
appB_in_gaa(x1, x2, x3)  =  appB_in_gaa(x1)
[]  =  []
appB_out_gaa(x1, x2, x3)  =  appB_out_gaa(x1)
.(x1, x2)  =  .(x1, x2)
U2_gaa(x1, x2, x3, x4, x5)  =  U2_gaa(x1, x2, x5)
prefixA_out_ga(x1, x2)  =  prefixA_out_ga(x1)
APPB_IN_GAA(x1, x2, x3)  =  APPB_IN_GAA(x1)

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:

APPB_IN_GAA(.(T19, T20), X31, .(T19, T22)) → APPB_IN_GAA(T20, X31, T22)

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

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:

APPB_IN_GAA(.(T19, T20)) → APPB_IN_GAA(T20)

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

(11) 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:

• APPB_IN_GAA(.(T19, T20)) → APPB_IN_GAA(T20)
  The graph contains the following edges 1 > 1

(12) YES