Termination w.r.t. Q proof of /tmp/tmpY15iQ1/14.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
and(x, true) → x
and(x, false) → false
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(0) → 0
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)
Log(x) → -(Log'(x), I(0))
Val(L(x)) → x
Val(N(x, l, r)) → x
Min(L(x)) → x
Min(N(x, l, r)) → Min(l)
Max(L(x)) → x
Max(N(x, l, r)) → Max(r)
BS(L(x)) → true
BS(N(x, l, r)) → and(and(ge(x, Max(l)), ge(Min(r), x)), and(BS(l), BS(r)))
Size(L(x)) → I(0)
Size(N(x, l, r)) → +(+(Size(l), Size(r)), I(1))
WB(L(x)) → true
WB(N(x, l, r)) → and(if(ge(Size(l), Size(r)), ge(I(0), -(Size(l), Size(r))), ge(I(0), -(Size(r), Size(l)))), and(WB(l), WB(r)))

Q is empty.

↳ QTRS

  ↳ RRRPoloQTRSProof

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
and(x, true) → x
and(x, false) → false
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(0) → 0
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)
Log(x) → -(Log'(x), I(0))
Val(L(x)) → x
Val(N(x, l, r)) → x
Min(L(x)) → x
Min(N(x, l, r)) → Min(l)
Max(L(x)) → x
Max(N(x, l, r)) → Max(r)
BS(L(x)) → true
BS(N(x, l, r)) → and(and(ge(x, Max(l)), ge(Min(r), x)), and(BS(l), BS(r)))
Size(L(x)) → I(0)
Size(N(x, l, r)) → +(+(Size(l), Size(r)), I(1))
WB(L(x)) → true
WB(N(x, l, r)) → and(if(ge(Size(l), Size(r)), ge(I(0), -(Size(l), Size(r))), ge(I(0), -(Size(r), Size(l)))), and(WB(l), WB(r)))

Q is empty.

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
and(x, true) → x
and(x, false) → false
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(0) → 0
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)
Log(x) → -(Log'(x), I(0))
Val(L(x)) → x
Val(N(x, l, r)) → x
Min(L(x)) → x
Min(N(x, l, r)) → Min(l)
Max(L(x)) → x
Max(N(x, l, r)) → Max(r)
BS(L(x)) → true
BS(N(x, l, r)) → and(and(ge(x, Max(l)), ge(Min(r), x)), and(BS(l), BS(r)))
Size(L(x)) → I(0)
Size(N(x, l, r)) → +(+(Size(l), Size(r)), I(1))
WB(L(x)) → true
WB(N(x, l, r)) → and(if(ge(Size(l), Size(r)), ge(I(0), -(Size(l), Size(r))), ge(I(0), -(Size(r), Size(l)))), and(WB(l), WB(r)))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

Val(L(x)) → x
Val(N(x, l, r)) → x
Min(L(x)) → x
Max(L(x)) → x
BS(L(x)) → true
Size(L(x)) → I(0)
WB(L(x)) → true

Used ordering:
Polynomial interpretation [25]:

POL(+(x₁, x₂)) = x₁ + x₂
POL(-(x₁, x₂)) = x₁ + x₂
POL(0) = 0
POL(1) = 0
POL(BS(x₁)) = 2·x₁
POL(I(x₁)) = 2·x₁
POL(L(x₁)) = 2 + x₁
POL(Log(x₁)) = 2·x₁
POL(Log'(x₁)) = 2·x₁
POL(Max(x₁)) = x₁
POL(Min(x₁)) = x₁
POL(N(x₁, x₂, x₃)) = 2·x₁ + x₂ + x₃
POL(O(x₁)) = 2·x₁
POL(Size(x₁)) = 2·x₁
POL(Val(x₁)) = 2 + 2·x₁
POL(WB(x₁)) = 2·x₁
POL(and(x₁, x₂)) = x₁ + 2·x₂
POL(false) = 0
POL(ge(x₁, x₂)) = x₁ + x₂
POL(if(x₁, x₂, x₃)) = x₁ + x₂ + 2·x₃
POL(l) = 0
POL(not(x₁)) = x₁
POL(r) = 0
POL(true) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
and(x, true) → x
and(x, false) → false
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(0) → 0
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)
Log(x) → -(Log'(x), I(0))
Min(N(x, l, r)) → Min(l)
Max(N(x, l, r)) → Max(r)
BS(N(x, l, r)) → and(and(ge(x, Max(l)), ge(Min(r), x)), and(BS(l), BS(r)))
Size(N(x, l, r)) → +(+(Size(l), Size(r)), I(1))
WB(N(x, l, r)) → and(if(ge(Size(l), Size(r)), ge(I(0), -(Size(l), Size(r))), ge(I(0), -(Size(r), Size(l)))), and(WB(l), WB(r)))

Q is empty.

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
and(x, true) → x
and(x, false) → false
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(0) → 0
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)
Log(x) → -(Log'(x), I(0))
Min(N(x, l, r)) → Min(l)
Max(N(x, l, r)) → Max(r)
BS(N(x, l, r)) → and(and(ge(x, Max(l)), ge(Min(r), x)), and(BS(l), BS(r)))
Size(N(x, l, r)) → +(+(Size(l), Size(r)), I(1))
WB(N(x, l, r)) → and(if(ge(Size(l), Size(r)), ge(I(0), -(Size(l), Size(r))), ge(I(0), -(Size(r), Size(l)))), and(WB(l), WB(r)))

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

Log(x) → -(Log'(x), I(0))
Min(N(x, l, r)) → Min(l)
Max(N(x, l, r)) → Max(r)
BS(N(x, l, r)) → and(and(ge(x, Max(l)), ge(Min(r), x)), and(BS(l), BS(r)))
Size(N(x, l, r)) → +(+(Size(l), Size(r)), I(1))
WB(N(x, l, r)) → and(if(ge(Size(l), Size(r)), ge(I(0), -(Size(l), Size(r))), ge(I(0), -(Size(r), Size(l)))), and(WB(l), WB(r)))

Used ordering:
Polynomial interpretation [25]:

POL(+(x₁, x₂)) = x₁ + 2·x₂
POL(-(x₁, x₂)) = x₁ + x₂
POL(0) = 0
POL(1) = 0
POL(BS(x₁)) = 2·x₁
POL(I(x₁)) = 2·x₁
POL(Log(x₁)) = 1 + 2·x₁
POL(Log'(x₁)) = 2·x₁
POL(Max(x₁)) = 1 + x₁
POL(Min(x₁)) = x₁
POL(N(x₁, x₂, x₃)) = 2 + 2·x₁ + x₂ + x₃
POL(O(x₁)) = 2·x₁
POL(Size(x₁)) = 2·x₁
POL(WB(x₁)) = 2·x₁
POL(and(x₁, x₂)) = x₁ + 2·x₂
POL(false) = 0
POL(ge(x₁, x₂)) = 2·x₁ + x₂
POL(if(x₁, x₂, x₃)) = x₁ + x₂ + 2·x₃
POL(l) = 0
POL(not(x₁)) = x₁
POL(r) = 0
POL(true) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
and(x, true) → x
and(x, false) → false
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(0) → 0
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

Q is empty.

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
and(x, true) → x
and(x, false) → false
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(0) → 0
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

and(x, true) → x
and(x, false) → false

Used ordering:
Polynomial interpretation [25]:

POL(+(x₁, x₂)) = x₁ + 2·x₂
POL(-(x₁, x₂)) = x₁ + 2·x₂
POL(0) = 0
POL(1) = 0
POL(I(x₁)) = 2·x₁
POL(Log'(x₁)) = 2·x₁
POL(O(x₁)) = 2·x₁
POL(and(x₁, x₂)) = 1 + x₁ + 2·x₂
POL(false) = 0
POL(ge(x₁, x₂)) = 2·x₁ + 2·x₂
POL(if(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(not(x₁)) = 2·x₁
POL(true) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(0) → 0
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

Q is empty.

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(0) → 0
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

Log'(0) → 0

Used ordering:
Polynomial interpretation [25]:

POL(+(x₁, x₂)) = x₁ + x₂
POL(-(x₁, x₂)) = x₁ + 2·x₂
POL(0) = 0
POL(1) = 0
POL(I(x₁)) = 2·x₁
POL(Log'(x₁)) = 2 + 2·x₁
POL(O(x₁)) = 2·x₁
POL(false) = 0
POL(ge(x₁, x₂)) = x₁ + x₂
POL(if(x₁, x₂, x₃)) = x₁ + x₂ + 2·x₃
POL(not(x₁)) = x₁
POL(true) = 0

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

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

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

+¹(I(x), O(y)) → +¹(x, y)
+¹(O(x), I(y)) → +¹(x, y)
+¹(x, +(y, z)) → +¹(x, y)
GE(O(x), I(y)) → NOT(ge(y, x))
-¹(I(x), I(y)) → O¹(-(x, y))
+¹(O(x), O(y)) → +¹(x, y)
-¹(O(x), I(y)) → -¹(-(x, y), I(1))
-¹(O(x), O(y)) → -¹(x, y)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
LOG'(O(x)) → IF(ge(x, I(0)), +(Log'(x), I(0)), 0)
LOG'(I(x)) → LOG'(x)
+¹(I(x), I(y)) → O¹(+(+(x, y), I(0)))
-¹(O(x), I(y)) → -¹(x, y)
-¹(I(x), O(y)) → -¹(x, y)
+¹(x, +(y, z)) → +¹(+(x, y), z)
LOG'(O(x)) → LOG'(x)
GE(O(x), I(y)) → GE(y, x)
GE(I(x), O(y)) → GE(x, y)
+¹(I(x), I(y)) → +¹(x, y)
LOG'(O(x)) → +¹(Log'(x), I(0))
-¹(I(x), I(y)) → -¹(x, y)
GE(I(x), I(y)) → GE(x, y)
GE(0, O(x)) → GE(0, x)
LOG'(I(x)) → +¹(Log'(x), I(0))
+¹(O(x), O(y)) → O¹(+(x, y))
GE(O(x), O(y)) → GE(x, y)
-¹(O(x), O(y)) → O¹(-(x, y))
LOG'(O(x)) → GE(x, I(0))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

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

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

+¹(I(x), O(y)) → +¹(x, y)
+¹(O(x), I(y)) → +¹(x, y)
+¹(x, +(y, z)) → +¹(x, y)
GE(O(x), I(y)) → NOT(ge(y, x))
-¹(I(x), I(y)) → O¹(-(x, y))
+¹(O(x), O(y)) → +¹(x, y)
-¹(O(x), I(y)) → -¹(-(x, y), I(1))
-¹(O(x), O(y)) → -¹(x, y)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
LOG'(O(x)) → IF(ge(x, I(0)), +(Log'(x), I(0)), 0)
LOG'(I(x)) → LOG'(x)
+¹(I(x), I(y)) → O¹(+(+(x, y), I(0)))
-¹(O(x), I(y)) → -¹(x, y)
-¹(I(x), O(y)) → -¹(x, y)
+¹(x, +(y, z)) → +¹(+(x, y), z)
LOG'(O(x)) → LOG'(x)
GE(O(x), I(y)) → GE(y, x)
GE(I(x), O(y)) → GE(x, y)
+¹(I(x), I(y)) → +¹(x, y)
LOG'(O(x)) → +¹(Log'(x), I(0))
-¹(I(x), I(y)) → -¹(x, y)
GE(I(x), I(y)) → GE(x, y)
GE(0, O(x)) → GE(0, x)
LOG'(I(x)) → +¹(Log'(x), I(0))
+¹(O(x), O(y)) → O¹(+(x, y))
GE(O(x), O(y)) → GE(x, y)
-¹(O(x), O(y)) → O¹(-(x, y))
LOG'(O(x)) → GE(x, I(0))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 9 less nodes.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

GE(0, O(x)) → GE(0, x)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

GE(0, O(x)) → GE(0, x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:

GE(0, O(x)) → GE(0, x)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

GE(I(x), I(y)) → GE(x, y)
GE(O(x), O(y)) → GE(x, y)
GE(I(x), O(y)) → GE(x, y)
GE(O(x), I(y)) → GE(y, x)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

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

GE(I(x), I(y)) → GE(x, y)
GE(O(x), O(y)) → GE(x, y)
GE(O(x), I(y)) → GE(y, x)
GE(I(x), O(y)) → GE(x, y)

From the DPs we obtained the following set of size-change graphs:

GE(I(x), I(y)) → GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2
GE(O(x), O(y)) → GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2
GE(I(x), O(y)) → GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2
GE(O(x), I(y)) → GE(y, x)
The graph contains the following edges 2 > 1, 1 > 2

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

                          ↳ QDP

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

-¹(I(x), I(y)) → -¹(x, y)
-¹(I(x), O(y)) → -¹(x, y)
-¹(O(x), I(y)) → -¹(x, y)
-¹(O(x), O(y)) → -¹(x, y)
-¹(O(x), I(y)) → -¹(-(x, y), I(1))

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                          ↳ QDP

                          ↳ QDP

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

-¹(O(x), I(y)) → -¹(x, y)
-¹(I(x), O(y)) → -¹(x, y)
-¹(I(x), I(y)) → -¹(x, y)
-¹(O(x), I(y)) → -¹(-(x, y), I(1))
-¹(O(x), O(y)) → -¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
O(0) → 0

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

-(x, 0) → x

Used ordering: POLO with Polynomial interpretation [25]:

POL(-(x₁, x₂)) = x₁ + x₂
POL(-¹(x₁, x₂)) = 2·x₁ + 2·x₂
POL(0) = 1
POL(1) = 0
POL(I(x₁)) = x₁
POL(O(x₁)) = x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

                          ↳ QDP

                          ↳ QDP

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

-¹(I(x), I(y)) → -¹(x, y)
-¹(I(x), O(y)) → -¹(x, y)
-¹(O(x), I(y)) → -¹(x, y)
-¹(O(x), O(y)) → -¹(x, y)
-¹(O(x), I(y)) → -¹(-(x, y), I(1))

The TRS R consists of the following rules:

-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
O(0) → 0

-¹(I(x), I(y)) → -¹(x, y)
-¹(I(x), O(y)) → -¹(x, y)
-¹(O(x), I(y)) → -¹(x, y)
-¹(O(x), O(y)) → -¹(x, y)
-¹(O(x), I(y)) → -¹(-(x, y), I(1))

Strictly oriented rules of the TRS R:

-(O(x), O(y)) → O(-(x, y))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
O(0) → 0

Used ordering: POLO with Polynomial interpretation [25]:

POL(-(x₁, x₂)) = x₁ + x₂
POL(-¹(x₁, x₂)) = 2·x₁ + 2·x₂
POL(0) = 0
POL(1) = 0
POL(I(x₁)) = 1 + x₁
POL(O(x₁)) = 1 + x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ RuleRemovalProof

                                      ↳ QDP

                                        ↳ PisEmptyProof

                          ↳ QDP

                          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

-(0, x) → 0
-(O(x), I(y)) → I(-(-(x, y), I(1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                          ↳ QDP

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

+¹(I(x), O(y)) → +¹(x, y)
+¹(O(x), I(y)) → +¹(x, y)
+¹(x, +(y, z)) → +¹(x, y)
+¹(x, +(y, z)) → +¹(+(x, y), z)
+¹(O(x), O(y)) → +¹(x, y)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
+¹(I(x), I(y)) → +¹(x, y)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                          ↳ QDP

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

+¹(O(x), I(y)) → +¹(x, y)
+¹(I(x), O(y)) → +¹(x, y)
+¹(x, +(y, z)) → +¹(x, y)
+¹(O(x), O(y)) → +¹(x, y)
+¹(x, +(y, z)) → +¹(+(x, y), z)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
+¹(I(x), I(y)) → +¹(x, y)

The TRS R consists of the following rules:

+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
O(0) → 0

+¹(O(x), I(y)) → +¹(x, y)
+¹(I(x), O(y)) → +¹(x, y)
+¹(x, +(y, z)) → +¹(x, y)
+¹(O(x), O(y)) → +¹(x, y)
+¹(x, +(y, z)) → +¹(+(x, y), z)
+¹(I(x), I(y)) → +¹(+(x, y), I(0))
+¹(I(x), I(y)) → +¹(x, y)

Strictly oriented rules of the TRS R:

+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
O(0) → 0

Used ordering: POLO with Polynomial interpretation [25]:

POL(+(x₁, x₂)) = 1 + x₁ + x₂
POL(+¹(x₁, x₂)) = x₁ + 2·x₂
POL(0) = 0
POL(I(x₁)) = 2 + x₁
POL(O(x₁)) = 1 + x₁

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ RuleRemovalProof

                                  ↳ QDP

                                    ↳ PisEmptyProof

                          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

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

LOG'(O(x)) → LOG'(x)
LOG'(I(x)) → LOG'(x)

The TRS R consists of the following rules:

O(0) → 0
+(0, x) → x
+(x, 0) → x
+(O(x), O(y)) → O(+(x, y))
+(O(x), I(y)) → I(+(x, y))
+(I(x), O(y)) → I(+(x, y))
+(I(x), I(y)) → O(+(+(x, y), I(0)))
+(x, +(y, z)) → +(+(x, y), z)
-(x, 0) → x
-(0, x) → 0
-(O(x), O(y)) → O(-(x, y))
-(O(x), I(y)) → I(-(-(x, y), I(1)))
-(I(x), O(y)) → I(-(x, y))
-(I(x), I(y)) → O(-(x, y))
not(true) → false
not(false) → true
if(true, x, y) → x
if(false, x, y) → y
ge(O(x), O(y)) → ge(x, y)
ge(O(x), I(y)) → not(ge(y, x))
ge(I(x), O(y)) → ge(x, y)
ge(I(x), I(y)) → ge(x, y)
ge(x, 0) → true
ge(0, O(x)) → ge(0, x)
ge(0, I(x)) → false
Log'(I(x)) → +(Log'(x), I(0))
Log'(O(x)) → if(ge(x, I(0)), +(Log'(x), I(0)), 0)

↳ QTRS

  ↳ RRRPoloQTRSProof

    ↳ QTRS

      ↳ RRRPoloQTRSProof

        ↳ QTRS

          ↳ RRRPoloQTRSProof

            ↳ QTRS

              ↳ RRRPoloQTRSProof

                ↳ QTRS

                  ↳ DependencyPairsProof

                    ↳ QDP

                      ↳ DependencyGraphProof

                        ↳ AND

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                          ↳ QDP

                            ↳ UsableRulesProof

                              ↳ QDP

                                ↳ QDPSizeChangeProof

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

LOG'(O(x)) → LOG'(x)
LOG'(I(x)) → LOG'(x)

From the DPs we obtained the following set of size-change graphs:

LOG'(O(x)) → LOG'(x)
The graph contains the following edges 1 > 1
LOG'(I(x)) → LOG'(x)
The graph contains the following edges 1 > 1