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:

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))

The TRS R 2 is

a → b
a → c

The signature Sigma is {c, a, b}

(2) Obligation:

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

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

(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:

PLUS(x, y) → PLUSITER(x, y, 0)
PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)
PLUSITER(x, y, z) → LE(x, z)
IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
LE(s(x), s(y)) → LE(x, y)
SUM(xs) → SUMITER(xs, 0)
SUMITER(xs, x) → IFSUM(isempty(xs), xs, x, plus(x, head(xs)))
SUMITER(xs, x) → ISEMPTY(xs)
SUMITER(xs, x) → PLUS(x, head(xs))
SUMITER(xs, x) → HEAD(xs)
IFSUM(false, xs, x, y) → SUMITER(tail(xs), y)
IFSUM(false, xs, x, y) → TAIL(xs)

The TRS R consists of the following rules:

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 7 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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:

LE(s(x), s(y)) → LE(x, y)

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

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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].

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

(11) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

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:

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

(13) TRUE

(14) Obligation:

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

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)

The TRS R consists of the following rules:

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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:

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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].

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

(18) Obligation:

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

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))

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

(19) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z) we obtained the following new rules [LPAR04]:

PLUSITER(z0, s(z1), s(z2)) → IFPLUS(le(z0, s(z2)), z0, s(z1), s(z2))

(20) Obligation:

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

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
PLUSITER(z0, s(z1), s(z2)) → IFPLUS(le(z0, s(z2)), z0, s(z1), s(z2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))

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

(21) 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 IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z)) the following chains were created:

• We consider the chain PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z), IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z)) which results in the following constraint:
  

  (1)    (IFPLUS(le(x3, x5), x3, x4, x5)=IFPLUS(false, x6, x7, x8) ⇒ IFPLUS(false, x6, x7, x8)≥PLUSITER(x6, s(x7), s(x8)))

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

  (2)    (le(x3, x5)=false ⇒ IFPLUS(false, x3, x4, x5)≥PLUSITER(x3, s(x4), s(x5)))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x3, x5)=false which results in the following new constraints:
  

  (3)    (false=false ⇒ IFPLUS(false, s(x18), x4, 0)≥PLUSITER(s(x18), s(x4), s(0)))

  
  

  (4)    (le(x21, x20)=false∧(∀x22:le(x21, x20)=false ⇒ IFPLUS(false, x21, x22, x20)≥PLUSITER(x21, s(x22), s(x20))) ⇒ IFPLUS(false, s(x21), x4, s(x20))≥PLUSITER(s(x21), s(x4), s(s(x20))))

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

  (5)    (IFPLUS(false, s(x18), x4, 0)≥PLUSITER(s(x18), s(x4), s(0)))

  
  
  We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (∀x22:le(x21, x20)=false ⇒ IFPLUS(false, x21, x22, x20)≥PLUSITER(x21, s(x22), s(x20))) with σ = [x22 / x4] which results in the following new constraint:
  

  (6)    (IFPLUS(false, x21, x4, x20)≥PLUSITER(x21, s(x4), s(x20)) ⇒ IFPLUS(false, s(x21), x4, s(x20))≥PLUSITER(s(x21), s(x4), s(s(x20))))

  
  
  

For Pair PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z) the following chains were created:

• We consider the chain IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z)), PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z) which results in the following constraint:
  

  (7)    (PLUSITER(x9, s(x10), s(x11))=PLUSITER(x12, x13, x14) ⇒ PLUSITER(x12, x13, x14)≥IFPLUS(le(x12, x14), x12, x13, x14))

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

  (8)    (PLUSITER(x9, s(x10), s(x11))≥IFPLUS(le(x9, s(x11)), x9, s(x10), s(x11)))

  
  
  

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

• IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))
  
  • (IFPLUS(false, s(x18), x4, 0)≥PLUSITER(s(x18), s(x4), s(0)))
    
  • (IFPLUS(false, x21, x4, x20)≥PLUSITER(x21, s(x4), s(x20)) ⇒ IFPLUS(false, s(x21), x4, s(x20))≥PLUSITER(s(x21), s(x4), s(s(x20))))
    
  
  
• PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)
  
  • (PLUSITER(x9, s(x10), s(x11))≥IFPLUS(le(x9, s(x11)), x9, s(x10), s(x11)))
    
  
  

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(IFPLUS(x₁, x₂, x₃, x₄)) = -1 - x₁ + x₂ - x₄   
POL(PLUSITER(x₁, x₂, x₃)) = -1 + x₁ - x₃   
POL(c) = -2   
POL(false) = 0   
POL(le(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))

The following pairs are in P_bound:

IFPLUS(false, x, y, z) → PLUSITER(x, s(y), s(z))

The following rules are usable:

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

(22) Obligation:

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

PLUSITER(x, y, z) → IFPLUS(le(x, z), x, y, z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))

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

(23) DependencyGraphProof (EQUIVALENT transformation)

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

(24) TRUE

(25) Obligation:

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

SUMITER(xs, x) → IFSUM(isempty(xs), xs, x, plus(x, head(xs)))
IFSUM(false, xs, x, y) → SUMITER(tail(xs), y)

The TRS R consists of the following rules:

plus(x, y) → plusIter(x, y, 0)
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
ifPlus(true, x, y, z) → y
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
sum(xs) → sumIter(xs, 0)
sumIter(xs, x) → ifSum(isempty(xs), xs, x, plus(x, head(xs)))
ifSum(true, xs, x, y) → x
ifSum(false, xs, x, y) → sumIter(tail(xs), y)
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
tail(nil) → nil
tail(cons(x, xs)) → xs
a → b
a → c

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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

(26) 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.

(27) Obligation:

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

SUMITER(xs, x) → IFSUM(isempty(xs), xs, x, plus(x, head(xs)))
IFSUM(false, xs, x, y) → SUMITER(tail(xs), y)

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, xs)) → xs
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
plus(x, y) → plusIter(x, y, 0)
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
ifPlus(true, x, y, z) → y

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))
a

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

(28) 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].

sum(x0)
sumIter(x0, x1)
ifSum(true, x0, x1, x2)
ifSum(false, x0, x1, x2)
a

(29) Obligation:

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

SUMITER(xs, x) → IFSUM(isempty(xs), xs, x, plus(x, head(xs)))
IFSUM(false, xs, x, y) → SUMITER(tail(xs), y)

The TRS R consists of the following rules:

tail(nil) → nil
tail(cons(x, xs)) → xs
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
plus(x, y) → plusIter(x, y, 0)
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
ifPlus(true, x, y, z) → y

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SUMITER(xs, x) → IFSUM(isempty(xs), xs, x, plus(x, head(xs)))
IFSUM(false, xs, x, y) → SUMITER(tail(xs), y)

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

POL(SUMITER(x₁, x₂)) = 13/4 + (5/2)x₁   
POL(IFSUM(x₁, x₂, x₃, x₄)) = 1/4 + (4)x₁ + (3/4)x₂   
POL(isempty(x₁)) = 1/2 + (1/4)x₁   
POL(plus(x₁, x₂)) = 5/2 + (7/2)x₁ + (1/2)x₂   
POL(head(x₁)) = (1/4)x₁   
POL(false) = 1   
POL(tail(x₁)) = 1/4 + (1/4)x₁   
POL(nil) = 0   
POL(error) = 15/4   
POL(cons(x₁, x₂)) = 9/4 + (4)x₁ + (4)x₂   
POL(true) = 1/2   
POL(ifPlus(x₁, x₂, x₃, x₄)) = 15/4 + (1/4)x₁ + (3/2)x₂ + (11/4)x₃ + (7/2)x₄   
POL(le(x₁, x₂)) = 4 + (5/2)x₂   
POL(0) = 0   
POL(s(x₁)) = (4)x₁   
POL(plusIter(x₁, x₂, x₃)) = 15/4 + (5/4)x₁ + (4)x₂ + (1/4)x₃   

The value of delta used in the strict ordering is 3/8.
The following usable rules [FROCOS05] were oriented:

isempty(nil) → true
isempty(cons(x, xs)) → false
tail(nil) → nil
tail(cons(x, xs)) → xs

(31) Obligation:

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

tail(nil) → nil
tail(cons(x, xs)) → xs
isempty(nil) → true
isempty(cons(x, xs)) → false
head(nil) → error
head(cons(x, xs)) → x
plus(x, y) → plusIter(x, y, 0)
ifPlus(false, x, y, z) → plusIter(x, s(y), s(z))
plusIter(x, y, z) → ifPlus(le(x, z), x, y, z)
le(s(x), 0) → false
le(0, y) → true
le(s(x), s(y)) → le(x, y)
ifPlus(true, x, y, z) → y

The set Q consists of the following terms:

plus(x0, x1)
plusIter(x0, x1, x2)
ifPlus(true, x0, x1, x2)
ifPlus(false, x0, x1, x2)
le(s(x0), 0)
le(0, x0)
le(s(x0), s(x1))
isempty(nil)
isempty(cons(x0, x1))
head(nil)
head(cons(x0, x1))
tail(nil)
tail(cons(x0, x1))

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE