Left Termination of the query pattern
prefix(a,g)
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(a,g)

(1) PrologToPiTRSViaGraphTransformerProof (SOUND transformation)

Transformed Prolog program to (Pi-)TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
2 [label="A: prefix(T1, T2)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
6 [label="6: prefix(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="B: app(T7, X6, T6)\n\nT6 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: app(T7, X6, T6), SCOPE: 2, CLAUSE: 1 | app(T7, X6, T6), SCOPE: 2, CLAUSE: 2\n\nT6 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: app(T7, X6, T6), SCOPE: 2, CLAUSE: 1\n\nT6 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: app(T7, X6, T6), SCOPE: 2, CLAUSE: 2\n\nT6 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];
30 [label="30: app(T22, X31, T21)\n\nT21 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
31 [label="31: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
32 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
2 -> 32 [arrowhead = none ];
32 -> 6;

33 [label="ONLY EVAL with clause\nprefix(X4, X5) :- app(X4, X6, X5).\nand substitutionT1 -> T7,\nX4 -> T7,\nT2 -> T6,\nX5 -> T6,\nT5 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 33 [arrowhead = none ];
33 -> 8;

34 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 34 [arrowhead = none ];
34 -> 12;

35 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
12 -> 35 [arrowhead = none ];
35 -> 14;

36 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
12 -> 36 [arrowhead = none ];
36 -> 15;

37 [label="EVAL with clause\napp([], X15, X15).\nand substitutionT7 -> [],\nX6 -> T12,\nX15 -> T12,\nT6 -> T12,\nX16 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
14 -> 37 [arrowhead = none ];
37 -> 19;

38 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
14 -> 38 [arrowhead = none ];
38 -> 20;

39 [label="EVAL with clause\napp(.(X27, X28), X29, .(X27, X30)) :- app(X28, X29, X30).\nand substitutionX27 -> T19,\nX28 -> T22,\nT7 -> .(T19, T22),\nX6 -> X31,\nX29 -> X31,\nX30 -> T21,\nT6 -> .(T19, T21),\nT20 -> T22", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 39 [arrowhead = none ];
39 -> 30;

40 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 40 [arrowhead = none ];
40 -> 31;

41 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
19 -> 41 [arrowhead = none ];
41 -> 21;

42 [label="INSTANCE with matching:\nT7 -> T22\nX6 -> X31\nT6 -> T21", fontsize=14, style = filled, fillcolor = lightgrey];
30 -> 42 [arrowhead = none , style=dashed];
42 -> 8[style=dashed];

}

(2) Obligation:

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

prefixA_in_ag(T7, T6) → U1_ag(T7, T6, appB_in_aag(T7, X6, T6))
appB_in_aag([], T12, T12) → appB_out_aag([], T12, T12)
appB_in_aag(.(T19, T22), X31, .(T19, T21)) → U2_aag(T19, T22, X31, T21, appB_in_aag(T22, X31, T21))
U2_aag(T19, T22, X31, T21, appB_out_aag(T22, X31, T21)) → appB_out_aag(.(T19, T22), X31, .(T19, T21))
U1_ag(T7, T6, appB_out_aag(T7, X6, T6)) → prefixA_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
prefixA_in_ag(x1, x2)  =  prefixA_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
appB_in_aag(x1, x2, x3)  =  appB_in_aag(x3)
appB_out_aag(x1, x2, x3)  =  appB_out_aag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x4, x5)
prefixA_out_ag(x1, x2)  =  prefixA_out_ag(x1, x2)

(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_AG(T7, T6) → U1_AG(T7, T6, appB_in_aag(T7, X6, T6))
PREFIXA_IN_AG(T7, T6) → APPB_IN_AAG(T7, X6, T6)
APPB_IN_AAG(.(T19, T22), X31, .(T19, T21)) → U2_AAG(T19, T22, X31, T21, appB_in_aag(T22, X31, T21))
APPB_IN_AAG(.(T19, T22), X31, .(T19, T21)) → APPB_IN_AAG(T22, X31, T21)

The TRS R consists of the following rules:

prefixA_in_ag(T7, T6) → U1_ag(T7, T6, appB_in_aag(T7, X6, T6))
appB_in_aag([], T12, T12) → appB_out_aag([], T12, T12)
appB_in_aag(.(T19, T22), X31, .(T19, T21)) → U2_aag(T19, T22, X31, T21, appB_in_aag(T22, X31, T21))
U2_aag(T19, T22, X31, T21, appB_out_aag(T22, X31, T21)) → appB_out_aag(.(T19, T22), X31, .(T19, T21))
U1_ag(T7, T6, appB_out_aag(T7, X6, T6)) → prefixA_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
prefixA_in_ag(x1, x2)  =  prefixA_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
appB_in_aag(x1, x2, x3)  =  appB_in_aag(x3)
appB_out_aag(x1, x2, x3)  =  appB_out_aag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x4, x5)
prefixA_out_ag(x1, x2)  =  prefixA_out_ag(x1, x2)
PREFIXA_IN_AG(x1, x2)  =  PREFIXA_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
APPB_IN_AAG(x1, x2, x3)  =  APPB_IN_AAG(x3)
U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x1, x4, 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_AG(T7, T6) → U1_AG(T7, T6, appB_in_aag(T7, X6, T6))
PREFIXA_IN_AG(T7, T6) → APPB_IN_AAG(T7, X6, T6)
APPB_IN_AAG(.(T19, T22), X31, .(T19, T21)) → U2_AAG(T19, T22, X31, T21, appB_in_aag(T22, X31, T21))
APPB_IN_AAG(.(T19, T22), X31, .(T19, T21)) → APPB_IN_AAG(T22, X31, T21)

The TRS R consists of the following rules:

prefixA_in_ag(T7, T6) → U1_ag(T7, T6, appB_in_aag(T7, X6, T6))
appB_in_aag([], T12, T12) → appB_out_aag([], T12, T12)
appB_in_aag(.(T19, T22), X31, .(T19, T21)) → U2_aag(T19, T22, X31, T21, appB_in_aag(T22, X31, T21))
U2_aag(T19, T22, X31, T21, appB_out_aag(T22, X31, T21)) → appB_out_aag(.(T19, T22), X31, .(T19, T21))
U1_ag(T7, T6, appB_out_aag(T7, X6, T6)) → prefixA_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
prefixA_in_ag(x1, x2)  =  prefixA_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
appB_in_aag(x1, x2, x3)  =  appB_in_aag(x3)
appB_out_aag(x1, x2, x3)  =  appB_out_aag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x4, x5)
prefixA_out_ag(x1, x2)  =  prefixA_out_ag(x1, x2)
PREFIXA_IN_AG(x1, x2)  =  PREFIXA_IN_AG(x2)
U1_AG(x1, x2, x3)  =  U1_AG(x2, x3)
APPB_IN_AAG(x1, x2, x3)  =  APPB_IN_AAG(x3)
U2_AAG(x1, x2, x3, x4, x5)  =  U2_AAG(x1, x4, 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_AAG(.(T19, T22), X31, .(T19, T21)) → APPB_IN_AAG(T22, X31, T21)

The TRS R consists of the following rules:

prefixA_in_ag(T7, T6) → U1_ag(T7, T6, appB_in_aag(T7, X6, T6))
appB_in_aag([], T12, T12) → appB_out_aag([], T12, T12)
appB_in_aag(.(T19, T22), X31, .(T19, T21)) → U2_aag(T19, T22, X31, T21, appB_in_aag(T22, X31, T21))
U2_aag(T19, T22, X31, T21, appB_out_aag(T22, X31, T21)) → appB_out_aag(.(T19, T22), X31, .(T19, T21))
U1_ag(T7, T6, appB_out_aag(T7, X6, T6)) → prefixA_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
prefixA_in_ag(x1, x2)  =  prefixA_in_ag(x2)
U1_ag(x1, x2, x3)  =  U1_ag(x2, x3)
appB_in_aag(x1, x2, x3)  =  appB_in_aag(x3)
appB_out_aag(x1, x2, x3)  =  appB_out_aag(x1, x2, x3)
.(x1, x2)  =  .(x1, x2)
U2_aag(x1, x2, x3, x4, x5)  =  U2_aag(x1, x4, x5)
prefixA_out_ag(x1, x2)  =  prefixA_out_ag(x1, x2)
APPB_IN_AAG(x1, x2, x3)  =  APPB_IN_AAG(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:

APPB_IN_AAG(.(T19, T22), X31, .(T19, T21)) → APPB_IN_AAG(T22, X31, T21)

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

APPB_IN_AAG(.(T19, T21)) → APPB_IN_AAG(T21)

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_AAG(.(T19, T21)) → APPB_IN_AAG(T21)
  The graph contains the following edges 1 > 1

(12) YES