Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(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:

GT(s(x), s(y)) → GT(x, y)
PLUS(s(x), y) → PLUS(x, y)
DOUBLE(s(x)) → DOUBLE(x)
AVERAGE(x, y) → AVER(plus(x, y), 0)
AVERAGE(x, y) → PLUS(x, y)
AVER(sum, z) → IF(gt(sum, double(z)), sum, z)
AVER(sum, z) → GT(sum, double(z))
AVER(sum, z) → DOUBLE(z)
IF(true, sum, z) → AVER(sum, s(z))

The TRS R consists of the following rules:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

The TRS R consists of the following rules:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

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

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(11) Obligation:

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

DOUBLE(s(x)) → DOUBLE(x)

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

(12) 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:

• DOUBLE(s(x)) → DOUBLE(x)
  The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

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

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(18) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

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

(19) 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:

• PLUS(s(x), y) → PLUS(x, y)
  The graph contains the following edges 1 > 1, 2 >= 2

(20) TRUE

(21) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

The TRS R consists of the following rules:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

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

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(25) Obligation:

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

GT(s(x), s(y)) → GT(x, y)

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

(26) 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:

• GT(s(x), s(y)) → GT(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(27) TRUE

(28) Obligation:

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

AVER(sum, z) → IF(gt(sum, double(z)), sum, z)
IF(true, sum, z) → AVER(sum, s(z))

The TRS R consists of the following rules:

gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
double(0) → 0
double(s(x)) → s(s(double(x)))
average(x, y) → aver(plus(x, y), 0)
aver(sum, z) → if(gt(sum, double(z)), sum, z)
if(true, sum, z) → aver(sum, s(z))
if(false, sum, z) → z

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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

(29) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(30) Obligation:

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

AVER(sum, z) → IF(gt(sum, double(z)), sum, z)
IF(true, sum, z) → AVER(sum, s(z))

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
double(0)
double(s(x0))
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

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

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

plus(0, x0)
plus(s(x0), x1)
average(x0, x1)
aver(x0, x1)
if(true, x0, x1)
if(false, x0, x1)

(32) Obligation:

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

AVER(sum, z) → IF(gt(sum, double(z)), sum, z)
IF(true, sum, z) → AVER(sum, s(z))

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
double(0)
double(s(x0))

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

(33) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule AVER(sum, z) → IF(gt(sum, double(z)), sum, z) we obtained the following new rules [LPAR04]:

AVER(z0, s(z1)) → IF(gt(z0, double(s(z1))), z0, s(z1))

(34) Obligation:

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

IF(true, sum, z) → AVER(sum, s(z))
AVER(z0, s(z1)) → IF(gt(z0, double(s(z1))), z0, s(z1))

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
double(0)
double(s(x0))

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

(35) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair AVER(sum, z) → IF(gt(sum, double(z)), sum, z) the following chains were created:

• We consider the chain AVER(sum, z) → IF(gt(sum, double(z)), sum, z), IF(true, sum, z) → AVER(sum, s(z)) which results in the following constraint:
  

  (1)    (IF(gt(x2, double(x3)), x2, x3)=IF(true, x4, x5) ⇒ AVER(x2, x3)≥IF(gt(x2, double(x3)), x2, x3))

  
  
  We simplified constraint (1) using rules (I), (II), (IV), (VII) which results in the following new constraint:
  

  (2)    (double(x3)=x12∧gt(x2, x12)=true ⇒ AVER(x2, x3)≥IF(gt(x2, double(x3)), x2, x3))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x2, x12)=true which results in the following new constraints:
  

  (3)    (true=true∧double(x3)=0 ⇒ AVER(s(x14), x3)≥IF(gt(s(x14), double(x3)), s(x14), x3))

  
  

  (4)    (gt(x16, x15)=true∧double(x3)=s(x15)∧(∀x17:gt(x16, x15)=true∧double(x17)=x15 ⇒ AVER(x16, x17)≥IF(gt(x16, double(x17)), x16, x17)) ⇒ AVER(s(x16), x3)≥IF(gt(s(x16), double(x3)), s(x16), x3))

  
  
  We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
  

  (5)    (double(x3)=0 ⇒ AVER(s(x14), x3)≥IF(gt(s(x14), double(x3)), s(x14), x3))

  
  
  We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on double(x3)=s(x15) which results in the following new constraint:
  

  (6)    (s(s(double(x19)))=s(x15)∧gt(x16, x15)=true∧(∀x17:gt(x16, x15)=true∧double(x17)=x15 ⇒ AVER(x16, x17)≥IF(gt(x16, double(x17)), x16, x17))∧(∀x20,x21,x22:double(x19)=s(x20)∧gt(x21, x20)=true∧(∀x22:gt(x21, x20)=true∧double(x22)=x20 ⇒ AVER(x21, x22)≥IF(gt(x21, double(x22)), x21, x22)) ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) ⇒ AVER(s(x16), s(x19))≥IF(gt(s(x16), double(s(x19))), s(x16), s(x19)))

  
  
  We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on double(x3)=0 which results in the following new constraint:
  

  (7)    (0=0 ⇒ AVER(s(x14), 0)≥IF(gt(s(x14), double(0)), s(x14), 0))

  
  
  We simplified constraint (7) using rules (I), (II) which results in the following new constraint:
  

  (8)    (AVER(s(x14), 0)≥IF(gt(s(x14), double(0)), s(x14), 0))

  
  
  We simplified constraint (6) using rules (I), (II), (IV) which results in the following new constraint:
  

  (9)    (s(double(x19))=x15∧gt(x16, x15)=true∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) ⇒ AVER(s(x16), s(x19))≥IF(gt(s(x16), double(s(x19))), s(x16), s(x19)))

  
  
  We simplified constraint (9) using rule (V) (with possible (I) afterwards) using induction on gt(x16, x15)=true which results in the following new constraints:
  

  (10)    (true=true∧s(double(x19))=0∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) ⇒ AVER(s(s(x24)), s(x19))≥IF(gt(s(s(x24)), double(s(x19))), s(s(x24)), s(x19)))

  
  

  (11)    (gt(x26, x25)=true∧s(double(x19))=s(x25)∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19))∧(∀x27,x28,x29:gt(x26, x25)=true∧s(double(x27))=x25∧(∀x28,x29:double(x27)=s(x28)∧gt(x29, x28)=true ⇒ AVER(s(x29), x27)≥IF(gt(s(x29), double(x27)), s(x29), x27)) ⇒ AVER(s(x26), s(x27))≥IF(gt(s(x26), double(s(x27))), s(x26), s(x27))) ⇒ AVER(s(s(x26)), s(x19))≥IF(gt(s(s(x26)), double(s(x19))), s(s(x26)), s(x19)))

  
  
  We solved constraint (10) using rules (I), (II).We simplified constraint (11) using rules (I), (II), (IV) which results in the following new constraint:
  

  (12)    (gt(x26, x25)=true∧double(x19)=x25∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) ⇒ AVER(s(s(x26)), s(x19))≥IF(gt(s(s(x26)), double(s(x19))), s(s(x26)), s(x19)))

  
  
  We simplified constraint (12) using rule (V) (with possible (I) afterwards) using induction on gt(x26, x25)=true which results in the following new constraints:
  

  (13)    (true=true∧double(x19)=0∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) ⇒ AVER(s(s(s(x31))), s(x19))≥IF(gt(s(s(s(x31))), double(s(x19))), s(s(s(x31))), s(x19)))

  
  

  (14)    (gt(x33, x32)=true∧double(x19)=s(x32)∧(∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19))∧(∀x34,x35,x36:gt(x33, x32)=true∧double(x34)=x32∧(∀x35,x36:double(x34)=s(x35)∧gt(x36, x35)=true ⇒ AVER(s(x36), x34)≥IF(gt(s(x36), double(x34)), s(x36), x34)) ⇒ AVER(s(s(x33)), s(x34))≥IF(gt(s(s(x33)), double(s(x34))), s(s(x33)), s(x34))) ⇒ AVER(s(s(s(x33))), s(x19))≥IF(gt(s(s(s(x33))), double(s(x19))), s(s(s(x33))), s(x19)))

  
  
  We simplified constraint (13) using rules (I), (II), (IV) which results in the following new constraint:
  

  (15)    (double(x19)=0 ⇒ AVER(s(s(s(x31))), s(x19))≥IF(gt(s(s(s(x31))), double(s(x19))), s(s(s(x31))), s(x19)))

  
  
  We simplified constraint (14) using rule (VI) where we applied the induction hypothesis (∀x20,x21:double(x19)=s(x20)∧gt(x21, x20)=true ⇒ AVER(s(x21), x19)≥IF(gt(s(x21), double(x19)), s(x21), x19)) with σ = [x20 / x32, x21 / x33] which results in the following new constraint:
  

  (16)    (AVER(s(x33), x19)≥IF(gt(s(x33), double(x19)), s(x33), x19)∧(∀x34,x35,x36:gt(x33, x32)=true∧double(x34)=x32∧(∀x35,x36:double(x34)=s(x35)∧gt(x36, x35)=true ⇒ AVER(s(x36), x34)≥IF(gt(s(x36), double(x34)), s(x36), x34)) ⇒ AVER(s(s(x33)), s(x34))≥IF(gt(s(s(x33)), double(s(x34))), s(s(x33)), s(x34))) ⇒ AVER(s(s(s(x33))), s(x19))≥IF(gt(s(s(s(x33))), double(s(x19))), s(s(s(x33))), s(x19)))

  
  
  We simplified constraint (15) using rule (V) (with possible (I) afterwards) using induction on double(x19)=0 which results in the following new constraint:
  

  (17)    (0=0 ⇒ AVER(s(s(s(x31))), s(0))≥IF(gt(s(s(s(x31))), double(s(0))), s(s(s(x31))), s(0)))

  
  
  We simplified constraint (17) using rules (I), (II) which results in the following new constraint:
  

  (18)    (AVER(s(s(s(x31))), s(0))≥IF(gt(s(s(s(x31))), double(s(0))), s(s(s(x31))), s(0)))

  
  
  We simplified constraint (16) using rule (IV) which results in the following new constraint:
  

  (19)    (AVER(s(x33), x19)≥IF(gt(s(x33), double(x19)), s(x33), x19) ⇒ AVER(s(s(s(x33))), s(x19))≥IF(gt(s(s(s(x33))), double(s(x19))), s(s(s(x33))), s(x19)))

  
  
  

For Pair IF(true, sum, z) → AVER(sum, s(z)) the following chains were created:

• We consider the chain IF(true, sum, z) → AVER(sum, s(z)), AVER(sum, z) → IF(gt(sum, double(z)), sum, z) which results in the following constraint:
  

  (20)    (AVER(x6, s(x7))=AVER(x8, x9) ⇒ IF(true, x6, x7)≥AVER(x6, s(x7)))

  
  
  We simplified constraint (20) using rules (I), (II), (IV) which results in the following new constraint:
  

  (21)    (IF(true, x6, x7)≥AVER(x6, s(x7)))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• AVER(sum, z) → IF(gt(sum, double(z)), sum, z)
  
  • (AVER(s(x14), 0)≥IF(gt(s(x14), double(0)), s(x14), 0))
    
  • (AVER(s(s(s(x31))), s(0))≥IF(gt(s(s(s(x31))), double(s(0))), s(s(s(x31))), s(0)))
    
  • (AVER(s(x33), x19)≥IF(gt(s(x33), double(x19)), s(x33), x19) ⇒ AVER(s(s(s(x33))), s(x19))≥IF(gt(s(s(s(x33))), double(s(x19))), s(s(s(x33))), s(x19)))
    
  
  
• IF(true, sum, z) → AVER(sum, s(z))
  
  • (IF(true, x6, x7)≥AVER(x6, s(x7)))
    
  
  

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(AVER(x₁, x₂)) = -1 + x₁ - x₂   
POL(IF(x₁, x₂, x₃)) = -1 - x₁ + x₂ - x₃   
POL(c) = -1   
POL(double(x₁)) = 0   
POL(false) = 0   
POL(gt(x₁, x₂)) = 0   
POL(s(x₁)) = 2 + x₁   
POL(true) = 0   

The following pairs are in P_>:

IF(true, sum, z) → AVER(sum, s(z))

The following pairs are in P_bound:

AVER(sum, z) → IF(gt(sum, double(z)), sum, z)

The following rules are usable:

true → gt(s(x), 0)
gt(x, y) → gt(s(x), s(y))
false → gt(0, y)

(36) Complex Obligation (AND)

(37) Obligation:

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

AVER(sum, z) → IF(gt(sum, double(z)), sum, z)

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
double(0)
double(s(x0))

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

(38) DependencyGraphProof (EQUIVALENT transformation)

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

(39) TRUE

(40) Obligation:

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

IF(true, sum, z) → AVER(sum, s(z))

The TRS R consists of the following rules:

double(0) → 0
double(s(x)) → s(s(double(x)))
gt(0, y) → false
gt(s(x), 0) → true
gt(s(x), s(y)) → gt(x, y)

The set Q consists of the following terms:

gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))
double(0)
double(s(x0))

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

(41) DependencyGraphProof (EQUIVALENT transformation)

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

(42) TRUE