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

(0) Obligation:

Clauses:

p(cons(X, nil)).
p(cons(s(s(X)), cons(Y, Xs))) :- ','(p(cons(X, cons(Y, Xs))), p(cons(s(s(s(s(Y)))), Xs))).
p(cons(0, Xs)) :- p(Xs).

Query: p(g)

(1) PrologToTRSTransformerProof (SOUND transformation)

Transformed Prolog program to TRS.

digraph dp_graph {
node [outthreshold=100, inthreshold=100];
3 [label="3: p(T1)\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow, color=red];
4 [label="4: p(T1), SCOPE: 1, CLAUSE: 0 | p(T1), SCOPE: 1, CLAUSE: 1 | p(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
7 [label="7: p(T1), SCOPE: 1, CLAUSE: 0\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
8 [label="8: p(T1), SCOPE: 1, CLAUSE: 1 | p(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
12 [label="12: true\n\n", fontsize=16, style=filled, fillcolor=lightyellow];
13 [label="13: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
15 [label="15: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
21 [label="21: p(T1), SCOPE: 1, CLAUSE: 1\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
22 [label="22: p(T1), SCOPE: 1, CLAUSE: 2\n\nT1 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
34 [label="34: ','(p(cons(T19, cons(T20, T21))), p(cons(s(s(s(s(T20)))), T21)))\n\nT19 is ground\nT20 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
36 [label="36: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
50 [label="50: p(cons(T19, cons(T20, T21)))\n\nT19 is ground\nT20 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
51 [label="51: p(cons(s(s(s(s(T20)))), T21))\n\nT20 is ground\nT21 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
60 [label="60: p(T30)\n\nT30 is ground", fontsize=16, style=filled, fillcolor=lightyellow];
61 [label="61: \n\n", fontsize=16, style=filled, fillcolor=lightyellow];
62 [label="CASE", fontsize=14, style = filled, fillcolor = lightcyan];
3 -> 62 [arrowhead = none ];
62 -> 4;

63 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 63 [arrowhead = none ];
63 -> 7;

64 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
4 -> 64 [arrowhead = none ];
64 -> 8;

65 [label="EVAL with clause\np(cons(X5, nil)).\nand substitutionX5 -> T6,\nT1 -> cons(T6, nil)", fontsize=14, style = filled, fillcolor = lightblue];
7 -> 65 [arrowhead = none ];
65 -> 12;

66 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
7 -> 66 [arrowhead = none ];
66 -> 13;

67 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 67 [arrowhead = none ];
67 -> 21;

68 [label="PARALLEL", fontsize=14, style = filled, fillcolor = violet];
8 -> 68 [arrowhead = none ];
68 -> 22;

69 [label="SUCCESS", fontsize=14, style = filled, fillcolor = lightgreen];
12 -> 69 [arrowhead = none ];
69 -> 15;

70 [label="EVAL with clause\np(cons(s(s(X18)), cons(X19, X20))) :- ','(p(cons(X18, cons(X19, X20))), p(cons(s(s(s(s(X19)))), X20))).\nand substitutionX18 -> T19,\nX19 -> T20,\nX20 -> T21,\nT1 -> cons(s(s(T19)), cons(T20, T21))", fontsize=14, style = filled, fillcolor = lightblue];
21 -> 70 [arrowhead = none ];
70 -> 34;

71 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
21 -> 71 [arrowhead = none ];
71 -> 36;

72 [label="EVAL with clause\np(cons(0, X29)) :- p(X29).\nand substitutionX29 -> T30,\nT1 -> cons(0, T30)", fontsize=14, style = filled, fillcolor = lightblue];
22 -> 72 [arrowhead = none ];
72 -> 60;

73 [label="EVAL-BACKTRACK", fontsize=14, style = filled, fillcolor = lightgreen];
22 -> 73 [arrowhead = none ];
73 -> 61;

74 [label="SPLIT 1", fontsize=14, style = filled, fillcolor = violet];
34 -> 74 [arrowhead = none ];
74 -> 50;

75 [label="SPLIT 2\nnew knowledge:\nT19 is ground\nT20 is ground\nT21 is ground", fontsize=14, style = filled, fillcolor = violet];
34 -> 75 [arrowhead = none ];
75 -> 51;

76 [label="INSTANCE with matching:\nT1 -> cons(T19, cons(T20, T21))", fontsize=14, style = filled, fillcolor = lightgrey];
50 -> 76 [arrowhead = none , style=dashed];
76 -> 3[style=dashed];

77 [label="INSTANCE with matching:\nT1 -> cons(s(s(s(s(T20)))), T21)", fontsize=14, style = filled, fillcolor = lightgrey];
51 -> 77 [arrowhead = none , style=dashed];
77 -> 3[style=dashed];

78 [label="INSTANCE with matching:\nT1 -> T30", fontsize=14, style = filled, fillcolor = lightgrey];
60 -> 78 [arrowhead = none , style=dashed];
78 -> 3[style=dashed];

}

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f3_in(cons(T6, nil)) → f3_out1
f3_in(cons(s(s(T19)), cons(T20, T21))) → U1(f34_in(T19, T20, T21), cons(s(s(T19)), cons(T20, T21)))
U1(f34_out1, cons(s(s(T19)), cons(T20, T21))) → f3_out1
f3_in(cons(0, T30)) → U2(f3_in(T30), cons(0, T30))
U2(f3_out1, cons(0, T30)) → f3_out1
f34_in(T19, T20, T21) → U3(f3_in(cons(T19, cons(T20, T21))), T19, T20, T21)
U3(f3_out1, T19, T20, T21) → U4(f3_in(cons(s(s(s(s(T20)))), T21)), T19, T20, T21)
U4(f3_out1, T19, T20, T21) → f34_out1

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

F3_IN(cons(s(s(T19)), cons(T20, T21))) → U1¹(f34_in(T19, T20, T21), cons(s(s(T19)), cons(T20, T21)))
F3_IN(cons(s(s(T19)), cons(T20, T21))) → F34_IN(T19, T20, T21)
F3_IN(cons(0, T30)) → U2¹(f3_in(T30), cons(0, T30))
F3_IN(cons(0, T30)) → F3_IN(T30)
F34_IN(T19, T20, T21) → U3¹(f3_in(cons(T19, cons(T20, T21))), T19, T20, T21)
F34_IN(T19, T20, T21) → F3_IN(cons(T19, cons(T20, T21)))
U3¹(f3_out1, T19, T20, T21) → U4¹(f3_in(cons(s(s(s(s(T20)))), T21)), T19, T20, T21)
U3¹(f3_out1, T19, T20, T21) → F3_IN(cons(s(s(s(s(T20)))), T21))

The TRS R consists of the following rules:

f3_in(cons(T6, nil)) → f3_out1
f3_in(cons(s(s(T19)), cons(T20, T21))) → U1(f34_in(T19, T20, T21), cons(s(s(T19)), cons(T20, T21)))
U1(f34_out1, cons(s(s(T19)), cons(T20, T21))) → f3_out1
f3_in(cons(0, T30)) → U2(f3_in(T30), cons(0, T30))
U2(f3_out1, cons(0, T30)) → f3_out1
f34_in(T19, T20, T21) → U3(f3_in(cons(T19, cons(T20, T21))), T19, T20, T21)
U3(f3_out1, T19, T20, T21) → U4(f3_in(cons(s(s(s(s(T20)))), T21)), T19, T20, T21)
U4(f3_out1, T19, T20, T21) → f34_out1

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 3 less nodes.

(6) Obligation:

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

F3_IN(cons(s(s(T19)), cons(T20, T21))) → F34_IN(T19, T20, T21)
F34_IN(T19, T20, T21) → U3¹(f3_in(cons(T19, cons(T20, T21))), T19, T20, T21)
U3¹(f3_out1, T19, T20, T21) → F3_IN(cons(s(s(s(s(T20)))), T21))
F34_IN(T19, T20, T21) → F3_IN(cons(T19, cons(T20, T21)))
F3_IN(cons(0, T30)) → F3_IN(T30)

The TRS R consists of the following rules:

f3_in(cons(T6, nil)) → f3_out1
f3_in(cons(s(s(T19)), cons(T20, T21))) → U1(f34_in(T19, T20, T21), cons(s(s(T19)), cons(T20, T21)))
U1(f34_out1, cons(s(s(T19)), cons(T20, T21))) → f3_out1
f3_in(cons(0, T30)) → U2(f3_in(T30), cons(0, T30))
U2(f3_out1, cons(0, T30)) → f3_out1
f34_in(T19, T20, T21) → U3(f3_in(cons(T19, cons(T20, T21))), T19, T20, T21)
U3(f3_out1, T19, T20, T21) → U4(f3_in(cons(s(s(s(s(T20)))), T21)), T19, T20, T21)
U4(f3_out1, T19, T20, T21) → f34_out1

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

F3_IN(cons(0, T30)) → F3_IN(T30)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(0) = 1   
POL(F34_IN(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(F3_IN(x₁)) = x₁   
POL(U1(x₁, x₂)) = 0   
POL(U2(x₁, x₂)) = 0   
POL(U3(x₁, x₂, x₃, x₄)) = 0   
POL(U3¹(x₁, x₂, x₃, x₄)) = x₂ + x₃ + x₄   
POL(U4(x₁, x₂, x₃, x₄)) = 0   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(f34_in(x₁, x₂, x₃)) = 0   
POL(f34_out1) = 0   
POL(f3_in(x₁)) = 1 + x₁   
POL(f3_out1) = 1   
POL(nil) = 1   
POL(s(x₁)) = x₁   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(8) Obligation:

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

F3_IN(cons(s(s(T19)), cons(T20, T21))) → F34_IN(T19, T20, T21)
F34_IN(T19, T20, T21) → U3¹(f3_in(cons(T19, cons(T20, T21))), T19, T20, T21)
U3¹(f3_out1, T19, T20, T21) → F3_IN(cons(s(s(s(s(T20)))), T21))
F34_IN(T19, T20, T21) → F3_IN(cons(T19, cons(T20, T21)))

The TRS R consists of the following rules:

f3_in(cons(T6, nil)) → f3_out1
f3_in(cons(s(s(T19)), cons(T20, T21))) → U1(f34_in(T19, T20, T21), cons(s(s(T19)), cons(T20, T21)))
U1(f34_out1, cons(s(s(T19)), cons(T20, T21))) → f3_out1
f3_in(cons(0, T30)) → U2(f3_in(T30), cons(0, T30))
U2(f3_out1, cons(0, T30)) → f3_out1
f34_in(T19, T20, T21) → U3(f3_in(cons(T19, cons(T20, T21))), T19, T20, T21)
U3(f3_out1, T19, T20, T21) → U4(f3_in(cons(s(s(s(s(T20)))), T21)), T19, T20, T21)
U4(f3_out1, T19, T20, T21) → f34_out1

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

U3¹(f3_out1, T19, T20, T21) → F3_IN(cons(s(s(s(s(T20)))), T21))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( U3¹(x₁, ..., x₄) ) = 2x₄ + 2

POL( f3_in(x₁) ) = max{0, -2}

POL( cons(x₁, x₂) ) = x₂ + 1

POL( s(x₁) ) = 0

POL( U1(x₁, x₂) ) = max{0, 2x₁ - 1}

POL( f34_in(x₁, ..., x₃) ) = x₃

POL( 0 ) = 2

POL( U2(x₁, x₂) ) = 1

POL( f34_out1 ) = 2

POL( f3_out1 ) = 0

POL( U3(x₁, ..., x₄) ) = max{0, x₂ + x₄ - 2}

POL( U4(x₁, ..., x₄) ) = max{0, x₁ - 1}

POL( nil ) = 0

POL( F3_IN(x₁) ) = max{0, 2x₁ - 2}

POL( F34_IN(x₁, ..., x₃) ) = 2x₃ + 2

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(10) Obligation:

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

F3_IN(cons(s(s(T19)), cons(T20, T21))) → F34_IN(T19, T20, T21)
F34_IN(T19, T20, T21) → U3¹(f3_in(cons(T19, cons(T20, T21))), T19, T20, T21)
F34_IN(T19, T20, T21) → F3_IN(cons(T19, cons(T20, T21)))

The TRS R consists of the following rules:

f3_in(cons(T6, nil)) → f3_out1
f3_in(cons(s(s(T19)), cons(T20, T21))) → U1(f34_in(T19, T20, T21), cons(s(s(T19)), cons(T20, T21)))
U1(f34_out1, cons(s(s(T19)), cons(T20, T21))) → f3_out1
f3_in(cons(0, T30)) → U2(f3_in(T30), cons(0, T30))
U2(f3_out1, cons(0, T30)) → f3_out1
f34_in(T19, T20, T21) → U3(f3_in(cons(T19, cons(T20, T21))), T19, T20, T21)
U3(f3_out1, T19, T20, T21) → U4(f3_in(cons(s(s(s(s(T20)))), T21)), T19, T20, T21)
U4(f3_out1, T19, T20, T21) → f34_out1

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(12) Obligation:

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

F34_IN(T19, T20, T21) → F3_IN(cons(T19, cons(T20, T21)))
F3_IN(cons(s(s(T19)), cons(T20, T21))) → F34_IN(T19, T20, T21)

The TRS R consists of the following rules:

f3_in(cons(T6, nil)) → f3_out1
f3_in(cons(s(s(T19)), cons(T20, T21))) → U1(f34_in(T19, T20, T21), cons(s(s(T19)), cons(T20, T21)))
U1(f34_out1, cons(s(s(T19)), cons(T20, T21))) → f3_out1
f3_in(cons(0, T30)) → U2(f3_in(T30), cons(0, T30))
U2(f3_out1, cons(0, T30)) → f3_out1
f34_in(T19, T20, T21) → U3(f3_in(cons(T19, cons(T20, T21))), T19, T20, T21)
U3(f3_out1, T19, T20, T21) → U4(f3_in(cons(s(s(s(s(T20)))), T21)), T19, T20, T21)
U4(f3_out1, T19, T20, T21) → f34_out1

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

(13) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(14) Obligation:

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

F34_IN(T19, T20, T21) → F3_IN(cons(T19, cons(T20, T21)))
F3_IN(cons(s(s(T19)), cons(T20, T21))) → F34_IN(T19, T20, T21)

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

(15) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F3_IN(cons(s(s(T19)), cons(T20, T21))) → F34_IN(T19, T20, T21)

No rules are removed from R.

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F34_IN(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(F3_IN(x₁)) = x₁   
POL(cons(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 2·x₁   

(16) Obligation:

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

F34_IN(T19, T20, T21) → F3_IN(cons(T19, cons(T20, T21)))

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

(17) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(18) TRUE