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

(0) Obligation:

Clauses:

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

Query: suffix(a,g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: suffix(T1, T2)\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: suffix(T1, T2), SCOPE: 1, CLAUSE: 0\n\nT2 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: app(X6, T7, T6)\n\nT6 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: app(X6, T7, T6), SCOPE: 2, CLAUSE: 1 | app(X6, T7, T6), SCOPE: 2, CLAUSE: 2\n\nT6 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: app(X6, T7, T6), SCOPE: 2, CLAUSE: 1\n\nT6 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: app(X6, T7, T6), SCOPE: 2, CLAUSE: 2\n\nT6 is ground\nX6 is free", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: app(X29, T22, T21)\n\nT21 is ground\nX29 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 12 [arrowhead = none ];
12 -> 2;

13 [label="ONLY EVAL with clause\nsuffix(X4, X5) :- app(X6, X4, X5).\nand substitutionT1 -> T7,\nX4 -> T7,\nT2 -> T6,\nX5 -> T6,\nT5 -> T7", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 13 [arrowhead = none ];
13 -> 3;

14 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 14 [arrowhead = none ];
14 -> 4;

15 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 15 [arrowhead = none ];
15 -> 5;

16 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 16 [arrowhead = none ];
16 -> 6;

17 [label="EVAL with clause\napp([], X11, X11).\nand substitutionX6 -> [],\nT7 -> T12,\nX11 -> T12,\nT6 -> T12", fontsize=14, style = filled, fillcolor = lightblue];
5 -> 17 [arrowhead = none ];
17 -> 7;

18 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 18 [arrowhead = none ];
18 -> 8;

19 [label="EVAL with clause\napp(.(X24, X25), X26, .(X24, X27)) :- app(X25, X26, X27).\nand substitutionX24 -> T20,\nX25 -> X29,\nX6 -> .(T20, X29),\nT7 -> T22,\nX26 -> T22,\nX28 -> T20,\nX27 -> T21,\nT6 -> .(T20, T21),\nT19 -> T22", fontsize=14, style = filled, fillcolor = lightblue];
6 -> 19 [arrowhead = none ];
19 -> 10;

20 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
6 -> 20 [arrowhead = none ];
20 -> 11;

21 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 21 [arrowhead = none ];
21 -> 9;

22 [label="INSTANCE with matching:\nX6 -> X29\nT7 -> T22\nT6 -> T21", fontsize=14, style = filled, fillcolor = lightgrey];
10 -> 22 [arrowhead = none , style=dashed];
22 -> 3[style=dashed];

}

(2) Obligation:

Clauses:

appA([], T12, T12).
appA(.(T20, X29), T22, .(T20, T21)) :- appA(X29, T22, T21).
suffixB(T7, T6) :- appA(X6, T7, T6).

Query: suffixB(a,g)

(3) PrologToPiTRSProof (SOUND transformation)

We use the technique of [TOCL09]. With regard to the inferred argument filtering the predicates were used in the following modes:
suffixB_in: (f,b)
appA_in: (f,f,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

suffixB_in_ag(T7, T6) → U2_ag(T7, T6, appA_in_aag(X6, T7, T6))
appA_in_aag([], T12, T12) → appA_out_aag([], T12, T12)
appA_in_aag(.(T20, X29), T22, .(T20, T21)) → U1_aag(T20, X29, T22, T21, appA_in_aag(X29, T22, T21))
U1_aag(T20, X29, T22, T21, appA_out_aag(X29, T22, T21)) → appA_out_aag(.(T20, X29), T22, .(T20, T21))
U2_ag(T7, T6, appA_out_aag(X6, T7, T6)) → suffixB_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
suffixB_in_ag(x1, x2)  =  suffixB_in_ag(x2)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
appA_in_aag(x1, x2, x3)  =  appA_in_aag(x3)
appA_out_aag(x1, x2, x3)  =  appA_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
suffixB_out_ag(x1, x2)  =  suffixB_out_ag(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:

suffixB_in_ag(T7, T6) → U2_ag(T7, T6, appA_in_aag(X6, T7, T6))
appA_in_aag([], T12, T12) → appA_out_aag([], T12, T12)
appA_in_aag(.(T20, X29), T22, .(T20, T21)) → U1_aag(T20, X29, T22, T21, appA_in_aag(X29, T22, T21))
U1_aag(T20, X29, T22, T21, appA_out_aag(X29, T22, T21)) → appA_out_aag(.(T20, X29), T22, .(T20, T21))
U2_ag(T7, T6, appA_out_aag(X6, T7, T6)) → suffixB_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
suffixB_in_ag(x1, x2)  =  suffixB_in_ag(x2)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
appA_in_aag(x1, x2, x3)  =  appA_in_aag(x3)
appA_out_aag(x1, x2, x3)  =  appA_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
suffixB_out_ag(x1, x2)  =  suffixB_out_ag(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:

SUFFIXB_IN_AG(T7, T6) → U2_AG(T7, T6, appA_in_aag(X6, T7, T6))
SUFFIXB_IN_AG(T7, T6) → APPA_IN_AAG(X6, T7, T6)
APPA_IN_AAG(.(T20, X29), T22, .(T20, T21)) → U1_AAG(T20, X29, T22, T21, appA_in_aag(X29, T22, T21))
APPA_IN_AAG(.(T20, X29), T22, .(T20, T21)) → APPA_IN_AAG(X29, T22, T21)

The TRS R consists of the following rules:

suffixB_in_ag(T7, T6) → U2_ag(T7, T6, appA_in_aag(X6, T7, T6))
appA_in_aag([], T12, T12) → appA_out_aag([], T12, T12)
appA_in_aag(.(T20, X29), T22, .(T20, T21)) → U1_aag(T20, X29, T22, T21, appA_in_aag(X29, T22, T21))
U1_aag(T20, X29, T22, T21, appA_out_aag(X29, T22, T21)) → appA_out_aag(.(T20, X29), T22, .(T20, T21))
U2_ag(T7, T6, appA_out_aag(X6, T7, T6)) → suffixB_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
suffixB_in_ag(x1, x2)  =  suffixB_in_ag(x2)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
appA_in_aag(x1, x2, x3)  =  appA_in_aag(x3)
appA_out_aag(x1, x2, x3)  =  appA_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
suffixB_out_ag(x1, x2)  =  suffixB_out_ag(x1)
SUFFIXB_IN_AG(x1, x2)  =  SUFFIXB_IN_AG(x2)
U2_AG(x1, x2, x3)  =  U2_AG(x3)
APPA_IN_AAG(x1, x2, x3)  =  APPA_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, x5)

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

(6) Obligation:

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

SUFFIXB_IN_AG(T7, T6) → U2_AG(T7, T6, appA_in_aag(X6, T7, T6))
SUFFIXB_IN_AG(T7, T6) → APPA_IN_AAG(X6, T7, T6)
APPA_IN_AAG(.(T20, X29), T22, .(T20, T21)) → U1_AAG(T20, X29, T22, T21, appA_in_aag(X29, T22, T21))
APPA_IN_AAG(.(T20, X29), T22, .(T20, T21)) → APPA_IN_AAG(X29, T22, T21)

The TRS R consists of the following rules:

suffixB_in_ag(T7, T6) → U2_ag(T7, T6, appA_in_aag(X6, T7, T6))
appA_in_aag([], T12, T12) → appA_out_aag([], T12, T12)
appA_in_aag(.(T20, X29), T22, .(T20, T21)) → U1_aag(T20, X29, T22, T21, appA_in_aag(X29, T22, T21))
U1_aag(T20, X29, T22, T21, appA_out_aag(X29, T22, T21)) → appA_out_aag(.(T20, X29), T22, .(T20, T21))
U2_ag(T7, T6, appA_out_aag(X6, T7, T6)) → suffixB_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
suffixB_in_ag(x1, x2)  =  suffixB_in_ag(x2)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
appA_in_aag(x1, x2, x3)  =  appA_in_aag(x3)
appA_out_aag(x1, x2, x3)  =  appA_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
suffixB_out_ag(x1, x2)  =  suffixB_out_ag(x1)
SUFFIXB_IN_AG(x1, x2)  =  SUFFIXB_IN_AG(x2)
U2_AG(x1, x2, x3)  =  U2_AG(x3)
APPA_IN_AAG(x1, x2, x3)  =  APPA_IN_AAG(x3)
U1_AAG(x1, x2, x3, x4, x5)  =  U1_AAG(x1, 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:

APPA_IN_AAG(.(T20, X29), T22, .(T20, T21)) → APPA_IN_AAG(X29, T22, T21)

The TRS R consists of the following rules:

suffixB_in_ag(T7, T6) → U2_ag(T7, T6, appA_in_aag(X6, T7, T6))
appA_in_aag([], T12, T12) → appA_out_aag([], T12, T12)
appA_in_aag(.(T20, X29), T22, .(T20, T21)) → U1_aag(T20, X29, T22, T21, appA_in_aag(X29, T22, T21))
U1_aag(T20, X29, T22, T21, appA_out_aag(X29, T22, T21)) → appA_out_aag(.(T20, X29), T22, .(T20, T21))
U2_ag(T7, T6, appA_out_aag(X6, T7, T6)) → suffixB_out_ag(T7, T6)

The argument filtering Pi contains the following mapping:
suffixB_in_ag(x1, x2)  =  suffixB_in_ag(x2)
U2_ag(x1, x2, x3)  =  U2_ag(x3)
appA_in_aag(x1, x2, x3)  =  appA_in_aag(x3)
appA_out_aag(x1, x2, x3)  =  appA_out_aag(x1, x2)
.(x1, x2)  =  .(x1, x2)
U1_aag(x1, x2, x3, x4, x5)  =  U1_aag(x1, x5)
suffixB_out_ag(x1, x2)  =  suffixB_out_ag(x1)
APPA_IN_AAG(x1, x2, x3)  =  APPA_IN_AAG(x3)

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:

APPA_IN_AAG(.(T20, X29), T22, .(T20, T21)) → APPA_IN_AAG(X29, T22, T21)

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

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:

APPA_IN_AAG(.(T20, T21)) → APPA_IN_AAG(T21)

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:

• APPA_IN_AAG(.(T20, T21)) → APPA_IN_AAG(T21)
  The graph contains the following edges 1 > 1

(14) YES