Left Termination proof of /tmp/tmpPq8F2G/pplus2.pl

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

0 Prolog

↳1 PrologToPrologProblemTransformerProof (⇒, 0 ms)

↳2 Prolog

↳3 PrologToPiTRSProof (⇒, 0 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 (⇒, 0 ms)

        ↳13 QDP

        ↳14 QDPSizeChangeProof (⇔, 0 ms)

        ↳15 YES

    ↳16 PiDP

        ↳17 UsableRulesProof (⇔, 1 ms)

        ↳18 PiDP

        ↳19 PiDPToQDPProof (⇒, 0 ms)

        ↳20 QDP

        ↳21 MRRProof (⇔, 24 ms)

        ↳22 QDP

        ↳23 PisEmptyProof (⇔, 0 ms)

        ↳24 YES

(0) Obligation:

Clauses:

p(s(0), 0).
p(s(s(X)), s(s(Y))) :- p(s(X), s(Y)).
plus(0, Y, Y).
plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z)).

Query: plus(g,a,a)

(1) PrologToPrologProblemTransformerProof (SOUND transformation)

Built Prolog problem from termination graph ICLP10.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
1 [label="1: plus(T1, T2, T3)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
2 [label="2: plus(T1, T2, T3), SCOPE: 1, CLAUSE: 2 | plus(T1, T2, T3), SCOPE: 1, CLAUSE: 3\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
3 [label="3: true | plus(0, T2, T3), SCOPE: 1, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
4 [label="4: plus(T1, T2, T3), SCOPE: 1, CLAUSE: 3\n\nT1 is ground\nX2 is free\nplus(T1, T2, T3) !=? plus(0, X2, X2)", fontsize=16, style=filled, fillcolor=lightyellow];
5 [label="5: plus(0, T2, T3), SCOPE: 1, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
6 [label="6: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: ','(p(s(T9), X12), plus(X12, T12, T13))\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
9 [label="9: ','(p(s(T9), X12), plus(X12, T12, T13)), SCOPE: 2, CLAUSE: 0 | ','(p(s(T9), X12), plus(X12, T12, T13)), SCOPE: 2, CLAUSE: 1\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
10 [label="10: ','(p(s(T9), X12), plus(X12, T12, T13)), SCOPE: 2, CLAUSE: 0\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
11 [label="11: ','(p(s(T9), X12), plus(X12, T12, T13)), SCOPE: 2, CLAUSE: 1\n\nT9 is ground\nX12 is free", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: plus(0, T12, T13)\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
14 [label="14: plus(0, T12, T13), SCOPE: 3, CLAUSE: 2 | plus(0, T12, T13), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: plus(0, T12, T13), SCOPE: 3, CLAUSE: 2\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
16 [label="16: plus(0, T12, T13), SCOPE: 3, CLAUSE: 3\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
17 [label="17: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
18 [label="18: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
19 [label="19: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
20 [label="20: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: ','(p(s(T23), s(X31)), plus(s(s(X31)), T12, T13))\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
23 [label="23: p(s(T23), s(X31))\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
24 [label="24: plus(s(s(T24)), T12, T13)\n\nT24 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
25 [label="25: p(s(T23), s(X31)), SCOPE: 4, CLAUSE: 0 | p(s(T23), s(X31)), SCOPE: 4, CLAUSE: 1\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
26 [label="26: p(s(T23), s(X31)), SCOPE: 4, CLAUSE: 1\n\nT23 is ground\nX31 is free", fontsize=16, style=filled, fillcolor=lightyellow];
27 [label="27: p(s(T27), s(X40))\n\nT27 is ground\nX40 is free", fontsize=16, style=filled, fillcolor=lightyellow];
28 [label="28: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
29 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
1 -> 29 [arrowhead = none ];
29 -> 2;

30 [label="EVAL with clause\nplus(0, X2, X2).\nand substitutionT1 -> 0,\nT2 -> T5,\nX2 -> T5,\nT3 -> T5", fontsize=14, style = filled, fillcolor = lightblue];
2 -> 30 [arrowhead = none ];
30 -> 3;

31 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
2 -> 31 [arrowhead = none ];
31 -> 4;

32 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
3 -> 32 [arrowhead = none ];
32 -> 5;

33 [label="EVAL with clause\nplus(s(X9), X10, s(X11)) :- ','(p(s(X9), X12), plus(X12, X10, X11)).\nand substitutionX9 -> T9,\nT1 -> s(T9),\nT2 -> T12,\nX10 -> T12,\nX11 -> T13,\nT3 -> s(T13),\nT10 -> T12,\nT11 -> T13", fontsize=14, style = filled, fillcolor = lightblue];
4 -> 33 [arrowhead = none ];
33 -> 7;

34 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
4 -> 34 [arrowhead = none ];
34 -> 8;

35 [label="BACKTRACK\nfor clause: plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
5 -> 35 [arrowhead = none ];
35 -> 6;

36 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
7 -> 36 [arrowhead = none ];
36 -> 9;

37 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 37 [arrowhead = none ];
37 -> 10;

38 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
9 -> 38 [arrowhead = none ];
38 -> 11;

39 [label="EVAL with clause\np(s(0), 0).\nand substitutionT9 -> 0,\nX12 -> 0", fontsize=14, style = filled, fillcolor = lightblue];
10 -> 39 [arrowhead = none ];
39 -> 12;

40 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
10 -> 40 [arrowhead = none ];
40 -> 13;

41 [label="EVAL with clause\np(s(s(X29)), s(s(X30))) :- p(s(X29), s(X30)).\nand substitutionX29 -> T23,\nT9 -> s(T23),\nX30 -> X31,\nX12 -> s(s(X31))", fontsize=14, style = filled, fillcolor = lightblue];
11 -> 41 [arrowhead = none ];
41 -> 21;

42 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
11 -> 42 [arrowhead = none ];
42 -> 22;

43 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
12 -> 43 [arrowhead = none ];
43 -> 14;

44 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 44 [arrowhead = none ];
44 -> 15;

45 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
14 -> 45 [arrowhead = none ];
45 -> 16;

46 [label="EVAL with clause\nplus(0, X19, X19).\nand substitutionT12 -> T20,\nX19 -> T20,\nT13 -> T20", fontsize=14, style = filled, fillcolor = lightblue];
15 -> 46 [arrowhead = none ];
46 -> 17;

47 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
15 -> 47 [arrowhead = none ];
47 -> 18;

48 [label="BACKTRACK\nfor clause: plus(s(X), Y, s(Z)) :- ','(p(s(X), U), plus(U, Y, Z))because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
16 -> 48 [arrowhead = none ];
48 -> 20;

49 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
17 -> 49 [arrowhead = none ];
49 -> 19;

50 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
21 -> 50 [arrowhead = none ];
50 -> 23;

51 [label="SPLIT 2\nnew knowledge:\nT23 is ground\nT24 is ground\nreplacements:X31 -> T24", fontsize=14, style = filled, fillcolor = violet];
21 -> 51 [arrowhead = none ];
51 -> 24;

52 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
23 -> 52 [arrowhead = none ];
52 -> 25;

53 [label="INSTANCE with matching:\nT1 -> s(s(T24))\nT2 -> T12\nT3 -> T13", fontsize=14, style = filled, fillcolor = lightgrey];
24 -> 53 [arrowhead = none , style=dashed];
53 -> 1[style=dashed];

54 [label="BACKTRACK\nfor clause: p(s(0), 0)because of non-unification", fontsize=14, style = filled, fillcolor = lightgreen];
25 -> 54 [arrowhead = none ];
54 -> 26;

55 [label="EVAL with clause\np(s(s(X38)), s(s(X39))) :- p(s(X38), s(X39)).\nand substitutionX38 -> T27,\nT23 -> s(T27),\nX39 -> X40,\nX31 -> s(X40)", fontsize=14, style = filled, fillcolor = lightblue];
26 -> 55 [arrowhead = none ];
55 -> 27;

56 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
26 -> 56 [arrowhead = none ];
56 -> 28;

57 [label="INSTANCE with matching:\nT23 -> T27\nX31 -> X40", fontsize=14, style = filled, fillcolor = lightgrey];
27 -> 57 [arrowhead = none , style=dashed];
57 -> 23[style=dashed];

}

(2) Obligation:

Clauses:

pA(s(T27), s(X40)) :- pA(T27, X40).
plusB(0, T5, T5).
plusB(s(0), T20, s(T20)).
plusB(s(s(T23)), T12, s(T13)) :- pA(T23, X31).
plusB(s(s(T23)), T12, s(T13)) :- ','(pA(T23, T24), plusB(s(s(T24)), T12, T13)).

Query: plusB(g,a,a)

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

plusB_in_gaa(0, T5, T5) → plusB_out_gaa(0, T5, T5)
plusB_in_gaa(s(0), T20, s(T20)) → plusB_out_gaa(s(0), T20, s(T20))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, pA_in_ga(T23, X31))
pA_in_ga(s(T27), s(X40)) → U1_ga(T27, X40, pA_in_ga(T27, X40))
U1_ga(T27, X40, pA_out_ga(T27, X40)) → pA_out_ga(s(T27), s(X40))
U2_gaa(T23, T12, T13, pA_out_ga(T23, X31)) → plusB_out_gaa(s(s(T23)), T12, s(T13))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, pA_in_ga(T23, T24))
U3_gaa(T23, T12, T13, pA_out_ga(T23, T24)) → U4_gaa(T23, T12, T13, plusB_in_gaa(s(s(T24)), T12, T13))
U4_gaa(T23, T12, T13, plusB_out_gaa(s(s(T24)), T12, T13)) → plusB_out_gaa(s(s(T23)), T12, s(T13))

Infinitary Constructor Rewriting Termination of PiTRS implies Termination of Prolog

(4) Obligation:

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

plusB_in_gaa(0, T5, T5) → plusB_out_gaa(0, T5, T5)
plusB_in_gaa(s(0), T20, s(T20)) → plusB_out_gaa(s(0), T20, s(T20))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, pA_in_ga(T23, X31))
pA_in_ga(s(T27), s(X40)) → U1_ga(T27, X40, pA_in_ga(T27, X40))
U1_ga(T27, X40, pA_out_ga(T27, X40)) → pA_out_ga(s(T27), s(X40))
U2_gaa(T23, T12, T13, pA_out_ga(T23, X31)) → plusB_out_gaa(s(s(T23)), T12, s(T13))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, pA_in_ga(T23, T24))
U3_gaa(T23, T12, T13, pA_out_ga(T23, T24)) → U4_gaa(T23, T12, T13, plusB_in_gaa(s(s(T24)), T12, T13))
U4_gaa(T23, T12, T13, plusB_out_gaa(s(s(T24)), T12, T13)) → plusB_out_gaa(s(s(T23)), T12, s(T13))

(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:

PLUSB_IN_GAA(s(s(T23)), T12, s(T13)) → U2_GAA(T23, T12, T13, pA_in_ga(T23, X31))
PLUSB_IN_GAA(s(s(T23)), T12, s(T13)) → PA_IN_GA(T23, X31)
PA_IN_GA(s(T27), s(X40)) → U1_GA(T27, X40, pA_in_ga(T27, X40))
PA_IN_GA(s(T27), s(X40)) → PA_IN_GA(T27, X40)
PLUSB_IN_GAA(s(s(T23)), T12, s(T13)) → U3_GAA(T23, T12, T13, pA_in_ga(T23, T24))
U3_GAA(T23, T12, T13, pA_out_ga(T23, T24)) → U4_GAA(T23, T12, T13, plusB_in_gaa(s(s(T24)), T12, T13))
U3_GAA(T23, T12, T13, pA_out_ga(T23, T24)) → PLUSB_IN_GAA(s(s(T24)), T12, T13)

The TRS R consists of the following rules:

plusB_in_gaa(0, T5, T5) → plusB_out_gaa(0, T5, T5)
plusB_in_gaa(s(0), T20, s(T20)) → plusB_out_gaa(s(0), T20, s(T20))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, pA_in_ga(T23, X31))
pA_in_ga(s(T27), s(X40)) → U1_ga(T27, X40, pA_in_ga(T27, X40))
U1_ga(T27, X40, pA_out_ga(T27, X40)) → pA_out_ga(s(T27), s(X40))
U2_gaa(T23, T12, T13, pA_out_ga(T23, X31)) → plusB_out_gaa(s(s(T23)), T12, s(T13))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, pA_in_ga(T23, T24))
U3_gaa(T23, T12, T13, pA_out_ga(T23, T24)) → U4_gaa(T23, T12, T13, plusB_in_gaa(s(s(T24)), T12, T13))
U4_gaa(T23, T12, T13, plusB_out_gaa(s(s(T24)), T12, T13)) → plusB_out_gaa(s(s(T23)), T12, s(T13))

The argument filtering Pi contains the following mapping:
plusB_in_gaa(x1, x2, x3) = plusB_in_gaa(x1)
0 = 0
plusB_out_gaa(x1, x2, x3) = plusB_out_gaa
s(x1) = s(x1)
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
pA_in_ga(x1, x2) = pA_in_ga(x1)
U1_ga(x1, x2, x3) = U1_ga(x3)
pA_out_ga(x1, x2) = pA_out_ga(x2)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
U4_gaa(x1, x2, x3, x4) = U4_gaa(x4)
PLUSB_IN_GAA(x1, x2, x3) = PLUSB_IN_GAA(x1)
U2_GAA(x1, x2, x3, x4) = U2_GAA(x4)
PA_IN_GA(x1, x2) = PA_IN_GA(x1)
U1_GA(x1, x2, x3) = U1_GA(x3)
U3_GAA(x1, x2, x3, x4) = U3_GAA(x4)
U4_GAA(x1, x2, x3, x4) = U4_GAA(x4)

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

(6) Obligation:

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

PLUSB_IN_GAA(s(s(T23)), T12, s(T13)) → U2_GAA(T23, T12, T13, pA_in_ga(T23, X31))
PLUSB_IN_GAA(s(s(T23)), T12, s(T13)) → PA_IN_GA(T23, X31)
PA_IN_GA(s(T27), s(X40)) → U1_GA(T27, X40, pA_in_ga(T27, X40))
PA_IN_GA(s(T27), s(X40)) → PA_IN_GA(T27, X40)
PLUSB_IN_GAA(s(s(T23)), T12, s(T13)) → U3_GAA(T23, T12, T13, pA_in_ga(T23, T24))
U3_GAA(T23, T12, T13, pA_out_ga(T23, T24)) → U4_GAA(T23, T12, T13, plusB_in_gaa(s(s(T24)), T12, T13))
U3_GAA(T23, T12, T13, pA_out_ga(T23, T24)) → PLUSB_IN_GAA(s(s(T24)), T12, T13)

The TRS R consists of the following rules:

plusB_in_gaa(0, T5, T5) → plusB_out_gaa(0, T5, T5)
plusB_in_gaa(s(0), T20, s(T20)) → plusB_out_gaa(s(0), T20, s(T20))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, pA_in_ga(T23, X31))
pA_in_ga(s(T27), s(X40)) → U1_ga(T27, X40, pA_in_ga(T27, X40))
U1_ga(T27, X40, pA_out_ga(T27, X40)) → pA_out_ga(s(T27), s(X40))
U2_gaa(T23, T12, T13, pA_out_ga(T23, X31)) → plusB_out_gaa(s(s(T23)), T12, s(T13))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, pA_in_ga(T23, T24))
U3_gaa(T23, T12, T13, pA_out_ga(T23, T24)) → U4_gaa(T23, T12, T13, plusB_in_gaa(s(s(T24)), T12, T13))
U4_gaa(T23, T12, T13, plusB_out_gaa(s(s(T24)), T12, T13)) → plusB_out_gaa(s(s(T23)), T12, s(T13))

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LOPSTR] contains 2 SCCs with 4 less nodes.

(8) Complex Obligation (AND)

(9) Obligation:

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

PA_IN_GA(s(T27), s(X40)) → PA_IN_GA(T27, X40)

The TRS R consists of the following rules:

plusB_in_gaa(0, T5, T5) → plusB_out_gaa(0, T5, T5)
plusB_in_gaa(s(0), T20, s(T20)) → plusB_out_gaa(s(0), T20, s(T20))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, pA_in_ga(T23, X31))
pA_in_ga(s(T27), s(X40)) → U1_ga(T27, X40, pA_in_ga(T27, X40))
U1_ga(T27, X40, pA_out_ga(T27, X40)) → pA_out_ga(s(T27), s(X40))
U2_gaa(T23, T12, T13, pA_out_ga(T23, X31)) → plusB_out_gaa(s(s(T23)), T12, s(T13))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, pA_in_ga(T23, T24))
U3_gaa(T23, T12, T13, pA_out_ga(T23, T24)) → U4_gaa(T23, T12, T13, plusB_in_gaa(s(s(T24)), T12, T13))
U4_gaa(T23, T12, T13, plusB_out_gaa(s(s(T24)), T12, T13)) → plusB_out_gaa(s(s(T23)), T12, s(T13))

The argument filtering Pi contains the following mapping:
plusB_in_gaa(x1, x2, x3) = plusB_in_gaa(x1)
0 = 0
plusB_out_gaa(x1, x2, x3) = plusB_out_gaa
s(x1) = s(x1)
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
pA_in_ga(x1, x2) = pA_in_ga(x1)
U1_ga(x1, x2, x3) = U1_ga(x3)
pA_out_ga(x1, x2) = pA_out_ga(x2)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
U4_gaa(x1, x2, x3, x4) = U4_gaa(x4)
PA_IN_GA(x1, x2) = PA_IN_GA(x1)

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:

PA_IN_GA(s(T27), s(X40)) → PA_IN_GA(T27, X40)

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

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

(12) PiDPToQDPProof (SOUND 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:

PA_IN_GA(s(T27)) → PA_IN_GA(T27)

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:

PA_IN_GA(s(T27)) → PA_IN_GA(T27)
The graph contains the following edges 1 > 1

(15) YES

(16) Obligation:

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

PLUSB_IN_GAA(s(s(T23)), T12, s(T13)) → U3_GAA(T23, T12, T13, pA_in_ga(T23, T24))
U3_GAA(T23, T12, T13, pA_out_ga(T23, T24)) → PLUSB_IN_GAA(s(s(T24)), T12, T13)

The TRS R consists of the following rules:

plusB_in_gaa(0, T5, T5) → plusB_out_gaa(0, T5, T5)
plusB_in_gaa(s(0), T20, s(T20)) → plusB_out_gaa(s(0), T20, s(T20))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U2_gaa(T23, T12, T13, pA_in_ga(T23, X31))
pA_in_ga(s(T27), s(X40)) → U1_ga(T27, X40, pA_in_ga(T27, X40))
U1_ga(T27, X40, pA_out_ga(T27, X40)) → pA_out_ga(s(T27), s(X40))
U2_gaa(T23, T12, T13, pA_out_ga(T23, X31)) → plusB_out_gaa(s(s(T23)), T12, s(T13))
plusB_in_gaa(s(s(T23)), T12, s(T13)) → U3_gaa(T23, T12, T13, pA_in_ga(T23, T24))
U3_gaa(T23, T12, T13, pA_out_ga(T23, T24)) → U4_gaa(T23, T12, T13, plusB_in_gaa(s(s(T24)), T12, T13))
U4_gaa(T23, T12, T13, plusB_out_gaa(s(s(T24)), T12, T13)) → plusB_out_gaa(s(s(T23)), T12, s(T13))

The argument filtering Pi contains the following mapping:
plusB_in_gaa(x1, x2, x3) = plusB_in_gaa(x1)
0 = 0
plusB_out_gaa(x1, x2, x3) = plusB_out_gaa
s(x1) = s(x1)
U2_gaa(x1, x2, x3, x4) = U2_gaa(x4)
pA_in_ga(x1, x2) = pA_in_ga(x1)
U1_ga(x1, x2, x3) = U1_ga(x3)
pA_out_ga(x1, x2) = pA_out_ga(x2)
U3_gaa(x1, x2, x3, x4) = U3_gaa(x4)
U4_gaa(x1, x2, x3, x4) = U4_gaa(x4)
PLUSB_IN_GAA(x1, x2, x3) = PLUSB_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4) = U3_GAA(x4)

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:

PLUSB_IN_GAA(s(s(T23)), T12, s(T13)) → U3_GAA(T23, T12, T13, pA_in_ga(T23, T24))
U3_GAA(T23, T12, T13, pA_out_ga(T23, T24)) → PLUSB_IN_GAA(s(s(T24)), T12, T13)

The TRS R consists of the following rules:

pA_in_ga(s(T27), s(X40)) → U1_ga(T27, X40, pA_in_ga(T27, X40))
U1_ga(T27, X40, pA_out_ga(T27, X40)) → pA_out_ga(s(T27), s(X40))

The argument filtering Pi contains the following mapping:
s(x1) = s(x1)
pA_in_ga(x1, x2) = pA_in_ga(x1)
U1_ga(x1, x2, x3) = U1_ga(x3)
pA_out_ga(x1, x2) = pA_out_ga(x2)
PLUSB_IN_GAA(x1, x2, x3) = PLUSB_IN_GAA(x1)
U3_GAA(x1, x2, x3, x4) = U3_GAA(x4)

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

(19) PiDPToQDPProof (SOUND 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:

PLUSB_IN_GAA(s(s(T23))) → U3_GAA(pA_in_ga(T23))
U3_GAA(pA_out_ga(T24)) → PLUSB_IN_GAA(s(s(T24)))

The TRS R consists of the following rules:

pA_in_ga(s(T27)) → U1_ga(pA_in_ga(T27))
U1_ga(pA_out_ga(X40)) → pA_out_ga(s(X40))

The set Q consists of the following terms:

pA_in_ga(x0)
U1_ga(x0)

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

(21) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

PLUSB_IN_GAA(s(s(T23))) → U3_GAA(pA_in_ga(T23))
U3_GAA(pA_out_ga(T24)) → PLUSB_IN_GAA(s(s(T24)))

Strictly oriented rules of the TRS R:

pA_in_ga(s(T27)) → U1_ga(pA_in_ga(T27))
U1_ga(pA_out_ga(X40)) → pA_out_ga(s(X40))

Used ordering: Knuth-Bendix order [KBO] with precedence:

pA_{in_ga₁} > U1_ga₁ > pA_{out_ga₁} > U3_GAA₁ > PLUSB_{IN_GAA₁} > s₁

and weight map:

pA_in_ga_1=1
s_1=3
U1_ga_1=3
pA_out_ga_1=2
PLUSB_IN_GAA_1=1
U3_GAA_1=5

The variable weight is 1

(22) Obligation:

Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

pA_in_ga(x0)
U1_ga(x0)

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

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(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 (SOUND transformation)

(13) Obligation:

(14) QDPSizeChangeProof (EQUIVALENT transformation)

(15) YES

(16) Obligation:

(17) UsableRulesProof (EQUIVALENT transformation)

(18) Obligation:

(19) PiDPToQDPProof (SOUND transformation)

(20) Obligation:

(21) MRRProof (EQUIVALENT transformation)

(22) Obligation:

(23) PisEmptyProof (EQUIVALENT transformation)

(24) YES