Left Termination proof of /tmp/tmphVBunG/ag01.pl

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

0 Prolog

↳1 PrologToPrologProblemTransformerProof (⇒, 0 ms)

↳2 Prolog

↳3 PrologToPiTRSProof (⇒, 4 ms)

↳4 PiTRS

↳5 DependencyPairsProof (⇔, 0 ms)

↳6 PiDP

↳7 DependencyGraphProof (⇔, 0 ms)

↳8 AND

    ↳9 PiDP

        ↳10 UsableRulesProof (⇔, 0 ms)

        ↳11 PiDP

        ↳12 PiDPToQDPProof (⇔, 1 ms)

        ↳13 QDP

        ↳14 QDPSizeChangeProof (⇔, 0 ms)

        ↳15 YES

    ↳16 PiDP

        ↳17 UsableRulesProof (⇔, 0 ms)

        ↳18 PiDP

        ↳19 PiDPToQDPProof (⇔, 0 ms)

        ↳20 QDP

        ↳21 QDPSizeChangeProof (⇔, 0 ms)

        ↳22 YES

(0) Obligation:

Clauses:

f(c(s(X), Y)) :- f(c(X, s(Y))).
g(c(X, s(Y))) :- g(c(s(X), Y)).
h(X) :- ','(f(X), g(X)).

Query: h(g)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: h(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: h(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: ','(f(T3), g(T3))\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: ','(f(T3), g(T3)), SCOPE: 2, CLAUSE: 0\n\nT3 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: ','(f(c(T8, s(T9))), g(c(s(T8), T9)))\n\nT8 is ground\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: f(c(T8, s(T9)))\n\nT8 is ground\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: g(c(s(T8), T9))\n\nT8 is ground\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: f(c(T8, s(T9))), SCOPE: 3, CLAUSE: 0\n\nT8 is ground\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: f(c(T18, s(s(T19))))\n\nT18 is ground\nT19 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: g(c(s(T8), T9)), SCOPE: 4, CLAUSE: 1\n\nT8 is ground\nT9 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: g(c(s(s(T28)), T29))\n\nT28 is ground\nT29 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 15 [arrowhead = none ];
15 -> 2;

16 [label="ONLY EVAL with clause\nh(X2) :- ','(f(X2), g(X2)).\nand substitutionT1 -> T3,\nX2 -> T3", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 16 [arrowhead = none ];
16 -> 3;

17 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 17 [arrowhead = none ];
17 -> 4;

18 [label="EVAL with clause\nf(c(s(X7), X8)) :- f(c(X7, s(X8))).\nand substitutionX7 -> T8,\nX8 -> T9,\nT3 -> c(s(T8), T9)", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 18 [arrowhead = none ];
18 -> 5;

19 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 19 [arrowhead = none ];
19 -> 6;

20 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
5 -> 20 [arrowhead = none ];
20 -> 7;

21 [label="SPLIT 2\nnew knowledge:\nT8 is ground\nT9 is ground", fontsize=14, style = filled, fillcolor = violet];
5 -> 21 [arrowhead = none ];
21 -> 8;

22 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 22 [arrowhead = none ];
22 -> 9;

23 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
8 -> 23 [arrowhead = none ];
23 -> 12;

24 [label="EVAL with clause\nf(c(s(X17), X18)) :- f(c(X17, s(X18))).\nand substitutionX17 -> T18,\nT8 -> s(T18),\nT9 -> T19,\nX18 -> s(T19)", fontsize=14, style = filled, fillcolor = lightblue];
9 -> 24 [arrowhead = none ];
24 -> 10;

25 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
9 -> 25 [arrowhead = none ];
25 -> 11;

26 [label="INSTANCE with matching:\nT8 -> T18\nT9 -> s(T19)", fontsize=14, style = filled, fillcolor = lightgrey];
10 -> 26 [arrowhead = none , style=dashed];
26 -> 7[style=dashed];

27 [label="EVAL with clause\ng(c(X27, s(X28))) :- g(c(s(X27), X28)).\nand substitutionT8 -> T28,\nX27 -> s(T28),\nX28 -> T29,\nT9 -> s(T29)", fontsize=14, style = filled, fillcolor = lightblue];
12 -> 27 [arrowhead = none ];
27 -> 13;

28 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 28 [arrowhead = none ];
28 -> 14;

29 [label="INSTANCE with matching:\nT8 -> s(T28)\nT9 -> T29", fontsize=14, style = filled, fillcolor = lightgrey];
13 -> 29 [arrowhead = none , style=dashed];
29 -> 8[style=dashed];

}

(2) Obligation:

Clauses:

fA(s(T18), T19) :- fA(T18, s(T19)).
gB(T28, s(T29)) :- gB(s(T28), T29).
hC(c(s(T8), T9)) :- fA(T8, T9).
hC(c(s(T8), T9)) :- ','(fA(T8, T9), gB(T8, T9)).

Query: hC(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:
hC_in: (b)
fA_in: (b,b)
gB_in: (b,b)
Transforming Prolog into the following Term Rewriting System:
Pi-finite rewrite system:
The TRS R consists of the following rules:

hC_in_g(c(s(T8), T9)) → U3_g(T8, T9, fA_in_gg(T8, T9))
fA_in_gg(s(T18), T19) → U1_gg(T18, T19, fA_in_gg(T18, s(T19)))
U1_gg(T18, T19, fA_out_gg(T18, s(T19))) → fA_out_gg(s(T18), T19)
U3_g(T8, T9, fA_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))
U3_g(T8, T9, fA_out_gg(T8, T9)) → U4_g(T8, T9, gB_in_gg(T8, T9))
gB_in_gg(T28, s(T29)) → U2_gg(T28, T29, gB_in_gg(s(T28), T29))
U2_gg(T28, T29, gB_out_gg(s(T28), T29)) → gB_out_gg(T28, s(T29))
U4_g(T8, T9, gB_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))

Pi is empty.

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

hC_in_g(c(s(T8), T9)) → U3_g(T8, T9, fA_in_gg(T8, T9))
fA_in_gg(s(T18), T19) → U1_gg(T18, T19, fA_in_gg(T18, s(T19)))
U1_gg(T18, T19, fA_out_gg(T18, s(T19))) → fA_out_gg(s(T18), T19)
U3_g(T8, T9, fA_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))
U3_g(T8, T9, fA_out_gg(T8, T9)) → U4_g(T8, T9, gB_in_gg(T8, T9))
gB_in_gg(T28, s(T29)) → U2_gg(T28, T29, gB_in_gg(s(T28), T29))
U2_gg(T28, T29, gB_out_gg(s(T28), T29)) → gB_out_gg(T28, s(T29))
U4_g(T8, T9, gB_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))

Pi is empty.

(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:

HC_IN_G(c(s(T8), T9)) → U3_G(T8, T9, fA_in_gg(T8, T9))
HC_IN_G(c(s(T8), T9)) → FA_IN_GG(T8, T9)
FA_IN_GG(s(T18), T19) → U1_GG(T18, T19, fA_in_gg(T18, s(T19)))
FA_IN_GG(s(T18), T19) → FA_IN_GG(T18, s(T19))
U3_G(T8, T9, fA_out_gg(T8, T9)) → U4_G(T8, T9, gB_in_gg(T8, T9))
U3_G(T8, T9, fA_out_gg(T8, T9)) → GB_IN_GG(T8, T9)
GB_IN_GG(T28, s(T29)) → U2_GG(T28, T29, gB_in_gg(s(T28), T29))
GB_IN_GG(T28, s(T29)) → GB_IN_GG(s(T28), T29)

The TRS R consists of the following rules:

hC_in_g(c(s(T8), T9)) → U3_g(T8, T9, fA_in_gg(T8, T9))
fA_in_gg(s(T18), T19) → U1_gg(T18, T19, fA_in_gg(T18, s(T19)))
U1_gg(T18, T19, fA_out_gg(T18, s(T19))) → fA_out_gg(s(T18), T19)
U3_g(T8, T9, fA_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))
U3_g(T8, T9, fA_out_gg(T8, T9)) → U4_g(T8, T9, gB_in_gg(T8, T9))
gB_in_gg(T28, s(T29)) → U2_gg(T28, T29, gB_in_gg(s(T28), T29))
U2_gg(T28, T29, gB_out_gg(s(T28), T29)) → gB_out_gg(T28, s(T29))
U4_g(T8, T9, gB_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))

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

(6) Obligation:

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

HC_IN_G(c(s(T8), T9)) → U3_G(T8, T9, fA_in_gg(T8, T9))
HC_IN_G(c(s(T8), T9)) → FA_IN_GG(T8, T9)
FA_IN_GG(s(T18), T19) → U1_GG(T18, T19, fA_in_gg(T18, s(T19)))
FA_IN_GG(s(T18), T19) → FA_IN_GG(T18, s(T19))
U3_G(T8, T9, fA_out_gg(T8, T9)) → U4_G(T8, T9, gB_in_gg(T8, T9))
U3_G(T8, T9, fA_out_gg(T8, T9)) → GB_IN_GG(T8, T9)
GB_IN_GG(T28, s(T29)) → U2_GG(T28, T29, gB_in_gg(s(T28), T29))
GB_IN_GG(T28, s(T29)) → GB_IN_GG(s(T28), T29)

The TRS R consists of the following rules:

hC_in_g(c(s(T8), T9)) → U3_g(T8, T9, fA_in_gg(T8, T9))
fA_in_gg(s(T18), T19) → U1_gg(T18, T19, fA_in_gg(T18, s(T19)))
U1_gg(T18, T19, fA_out_gg(T18, s(T19))) → fA_out_gg(s(T18), T19)
U3_g(T8, T9, fA_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))
U3_g(T8, T9, fA_out_gg(T8, T9)) → U4_g(T8, T9, gB_in_gg(T8, T9))
gB_in_gg(T28, s(T29)) → U2_gg(T28, T29, gB_in_gg(s(T28), T29))
U2_gg(T28, T29, gB_out_gg(s(T28), T29)) → gB_out_gg(T28, s(T29))
U4_g(T8, T9, gB_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 6 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

GB_IN_GG(T28, s(T29)) → GB_IN_GG(s(T28), T29)

The TRS R consists of the following rules:

hC_in_g(c(s(T8), T9)) → U3_g(T8, T9, fA_in_gg(T8, T9))
fA_in_gg(s(T18), T19) → U1_gg(T18, T19, fA_in_gg(T18, s(T19)))
U1_gg(T18, T19, fA_out_gg(T18, s(T19))) → fA_out_gg(s(T18), T19)
U3_g(T8, T9, fA_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))
U3_g(T8, T9, fA_out_gg(T8, T9)) → U4_g(T8, T9, gB_in_gg(T8, T9))
gB_in_gg(T28, s(T29)) → U2_gg(T28, T29, gB_in_gg(s(T28), T29))
U2_gg(T28, T29, gB_out_gg(s(T28), T29)) → gB_out_gg(T28, s(T29))
U4_g(T8, T9, gB_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))

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

(10) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(11) Obligation:

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

GB_IN_GG(T28, s(T29)) → GB_IN_GG(s(T28), T29)

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

(12) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(13) Obligation:

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

GB_IN_GG(T28, s(T29)) → GB_IN_GG(s(T28), T29)

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

(14) 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:

GB_IN_GG(T28, s(T29)) → GB_IN_GG(s(T28), T29)
The graph contains the following edges 2 > 2

(15) YES

(16) Obligation:

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

FA_IN_GG(s(T18), T19) → FA_IN_GG(T18, s(T19))

The TRS R consists of the following rules:

hC_in_g(c(s(T8), T9)) → U3_g(T8, T9, fA_in_gg(T8, T9))
fA_in_gg(s(T18), T19) → U1_gg(T18, T19, fA_in_gg(T18, s(T19)))
U1_gg(T18, T19, fA_out_gg(T18, s(T19))) → fA_out_gg(s(T18), T19)
U3_g(T8, T9, fA_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))
U3_g(T8, T9, fA_out_gg(T8, T9)) → U4_g(T8, T9, gB_in_gg(T8, T9))
gB_in_gg(T28, s(T29)) → U2_gg(T28, T29, gB_in_gg(s(T28), T29))
U2_gg(T28, T29, gB_out_gg(s(T28), T29)) → gB_out_gg(T28, s(T29))
U4_g(T8, T9, gB_out_gg(T8, T9)) → hC_out_g(c(s(T8), T9))

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

(17) UsableRulesProof (EQUIVALENT transformation)

For (infinitary) constructor rewriting [LOPSTR] we can delete all non-usable rules from R.

(18) Obligation:

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

FA_IN_GG(s(T18), T19) → FA_IN_GG(T18, s(T19))

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

(19) PiDPToQDPProof (EQUIVALENT transformation)

Transforming (infinitary) constructor rewriting Pi-DP problem [LOPSTR] into ordinary QDP problem [LPAR04] by application of Pi.

(20) Obligation:

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

FA_IN_GG(s(T18), T19) → FA_IN_GG(T18, s(T19))

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

(21) 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:

FA_IN_GG(s(T18), T19) → FA_IN_GG(T18, s(T19))
The graph contains the following edges 1 > 1

(0) Obligation:

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

(2) Obligation:

(3) PrologToPiTRSProof (SOUND transformation)

(4) Obligation:

(5) DependencyPairsProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Complex Obligation (AND)

(9) Obligation:

(10) UsableRulesProof (EQUIVALENT transformation)

(11) Obligation:

(12) PiDPToQDPProof (EQUIVALENT transformation)

(13) Obligation:

(14) QDPSizeChangeProof (EQUIVALENT transformation)

(15) YES

(16) Obligation:

(17) UsableRulesProof (EQUIVALENT transformation)

(18) Obligation:

(19) PiDPToQDPProof (EQUIVALENT transformation)

(20) Obligation:

(21) QDPSizeChangeProof (EQUIVALENT transformation)

(22) YES