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:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

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:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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:

DIV(x, s(y)) → D(x, s(y), 0)
D(x, s(y), z) → COND(ge(x, z), x, y, z)
D(x, s(y), z) → GE(x, z)
COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
COND(true, x, y, z) → PLUS(s(y), z)
GE(s(u), s(v)) → GE(u, v)
PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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 3 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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:

PLUS(n, s(m)) → PLUS(n, m)

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

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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].

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

(11) Obligation:

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

PLUS(n, s(m)) → PLUS(n, m)

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:

• PLUS(n, s(m)) → PLUS(n, m)
  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:

GE(s(u), s(v)) → GE(u, v)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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:

GE(s(u), s(v)) → GE(u, v)

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

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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].

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

(18) Obligation:

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

GE(s(u), s(v)) → GE(u, v)

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:

• GE(s(u), s(v)) → GE(u, v)
  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:

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

div(x, s(y)) → d(x, s(y), 0)
d(x, s(y), z) → cond(ge(x, z), x, y, z)
cond(true, x, y, z) → s(d(x, s(y), plus(s(y), z)))
cond(false, x, y, z) → 0
ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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:

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)
ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(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].

div(x0, s(x1))
d(x0, s(x1), x2)
cond(true, x0, x1, x2)
cond(false, x0, x1, x2)

(25) Obligation:

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

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

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

(26) 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 COND(true, x, y, z) → D(x, s(y), plus(s(y), z)) the following chains were created:

• We consider the chain D(x, s(y), z) → COND(ge(x, z), x, y, z), COND(true, x, y, z) → D(x, s(y), plus(s(y), z)) which results in the following constraint:
  

  (1)    (COND(ge(x3, x5), x3, x4, x5)=COND(true, x6, x7, x8) ⇒ COND(true, x6, x7, x8)≥D(x6, s(x7), plus(s(x7), x8)))

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

  (2)    (ge(x3, x5)=true ⇒ COND(true, x3, x4, x5)≥D(x3, s(x4), plus(s(x4), x5)))

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

  (3)    (true=true ⇒ COND(true, x18, x4, 0)≥D(x18, s(x4), plus(s(x4), 0)))

  
  

  (4)    (ge(x21, x20)=true∧(∀x22:ge(x21, x20)=true ⇒ COND(true, x21, x22, x20)≥D(x21, s(x22), plus(s(x22), x20))) ⇒ COND(true, s(x21), x4, s(x20))≥D(s(x21), s(x4), plus(s(x4), s(x20))))

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

  (5)    (COND(true, x18, x4, 0)≥D(x18, s(x4), plus(s(x4), 0)))

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

  (6)    (COND(true, x21, x4, x20)≥D(x21, s(x4), plus(s(x4), x20)) ⇒ COND(true, s(x21), x4, s(x20))≥D(s(x21), s(x4), plus(s(x4), s(x20))))

  
  
  

For Pair D(x, s(y), z) → COND(ge(x, z), x, y, z) the following chains were created:

• We consider the chain COND(true, x, y, z) → D(x, s(y), plus(s(y), z)), D(x, s(y), z) → COND(ge(x, z), x, y, z) which results in the following constraint:
  

  (7)    (D(x9, s(x10), plus(s(x10), x11))=D(x12, s(x13), x14) ⇒ D(x12, s(x13), x14)≥COND(ge(x12, x14), x12, x13, x14))

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

  (8)    (D(x9, s(x10), x14)≥COND(ge(x9, x14), x9, x10, x14))

  
  
  

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

• COND(true, x, y, z) → D(x, s(y), plus(s(y), z))
  
  • (COND(true, x18, x4, 0)≥D(x18, s(x4), plus(s(x4), 0)))
    
  • (COND(true, x21, x4, x20)≥D(x21, s(x4), plus(s(x4), x20)) ⇒ COND(true, s(x21), x4, s(x20))≥D(s(x21), s(x4), plus(s(x4), s(x20))))
    
  
  
• D(x, s(y), z) → COND(ge(x, z), x, y, z)
  
  • (D(x9, s(x10), x14)≥COND(ge(x9, x14), x9, x10, x14))
    
  
  

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(COND(x₁, x₂, x₃, x₄)) = -1 - x₁ + x₂ - x₄   
POL(D(x₁, x₂, x₃)) = -1 + x₁ + x₂ - x₃   
POL(c) = -1   
POL(false) = 0   
POL(ge(x₁, x₂)) = 0   
POL(plus(x₁, x₂)) = x₁ + x₂   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

D(x, s(y), z) → COND(ge(x, z), x, y, z)

The following pairs are in P_bound:

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))

The following rules are usable:

s(plus(n, m)) → plus(n, s(m))
true → ge(u, 0)
false → ge(0, s(v))
ge(u, v) → ge(s(u), s(v))
n → plus(n, 0)

(27) Complex Obligation (AND)

(28) Obligation:

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

COND(true, x, y, z) → D(x, s(y), plus(s(y), z))

The TRS R consists of the following rules:

ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

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

(29) DependencyGraphProof (EQUIVALENT transformation)

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

(30) TRUE

(31) Obligation:

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

D(x, s(y), z) → COND(ge(x, z), x, y, z)

The TRS R consists of the following rules:

ge(u, 0) → true
ge(0, s(v)) → false
ge(s(u), s(v)) → ge(u, v)
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))

The set Q consists of the following terms:

ge(x0, 0)
ge(0, s(x0))
ge(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))

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

(32) DependencyGraphProof (EQUIVALENT transformation)

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

(33) TRUE