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:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

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:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

(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:

LT(s(x), s(y)) → LT(x, y)
PLUS(x, s(y)) → PLUS(x, y)
QUOT(x, s(y)) → HELP(x, s(y), 0)
HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
HELP(x, s(y), c) → LT(c, x)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))
IF(true, x, s(y), c) → PLUS(c, s(y))

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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(x, s(y)) → PLUS(x, y)

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

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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].

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

(11) Obligation:

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

PLUS(x, s(y)) → PLUS(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:

• PLUS(x, s(y)) → PLUS(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:

LT(s(x), s(y)) → LT(x, y)

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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:

LT(s(x), s(y)) → LT(x, y)

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

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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].

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

(18) Obligation:

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

LT(s(x), s(y)) → LT(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:

• LT(s(x), s(y)) → LT(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:

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))

The TRS R consists of the following rules:

lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)
plus(x, 0) → x
plus(x, s(y)) → s(plus(x, y))
quot(x, s(y)) → help(x, s(y), 0)
help(x, s(y), c) → if(lt(c, x), x, s(y), c)
if(true, x, s(y), c) → s(help(x, s(y), plus(c, s(y))))
if(false, x, s(y), c) → 0

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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:

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))

The TRS R consists of the following rules:

plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(s(x0), s(x1))
plus(x0, 0)
plus(x0, s(x1))
quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

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].

quot(x0, s(x1))
help(x0, s(x1), x2)
if(true, x0, s(x1), x2)
if(false, x0, s(x1), x2)

(25) Obligation:

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

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y)))

The TRS R consists of the following rules:

plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

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

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

(26) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule IF(true, x, s(y), c) → HELP(x, s(y), plus(c, s(y))) at position [2] we obtained the following new rules [LPAR04]:

IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))

(27) Obligation:

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

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))

The TRS R consists of the following rules:

plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

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

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

(28) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c) we obtained the following new rules [LPAR04]:

HELP(z0, s(z1), s(y_0)) → IF(lt(s(y_0), z0), z0, s(z1), s(y_0))

(29) Obligation:

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

IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))
HELP(z0, s(z1), s(y_0)) → IF(lt(s(y_0), z0), z0, s(z1), s(y_0))

The TRS R consists of the following rules:

plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

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

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

(30) 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 HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c) the following chains were created:

• We consider the chain IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y))), HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c) which results in the following constraint:
  

  (1)    (HELP(x3, s(x4), s(plus(x5, x4)))=HELP(x6, s(x7), x8) ⇒ HELP(x6, s(x7), x8)≥IF(lt(x8, x6), x6, s(x7), x8))

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

  (2)    (HELP(x3, s(x4), x8)≥IF(lt(x8, x3), x3, s(x4), x8))

  
  
  

For Pair IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y))) the following chains were created:

• We consider the chain HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c), IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y))) which results in the following constraint:
  

  (3)    (IF(lt(x11, x9), x9, s(x10), x11)=IF(true, x12, s(x13), x14) ⇒ IF(true, x12, s(x13), x14)≥HELP(x12, s(x13), s(plus(x14, x13))))

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

  (4)    (lt(x11, x9)=true ⇒ IF(true, x9, s(x10), x11)≥HELP(x9, s(x10), s(plus(x11, x10))))

  
  
  We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on lt(x11, x9)=true which results in the following new constraints:
  

  (5)    (true=true ⇒ IF(true, s(x19), s(x10), 0)≥HELP(s(x19), s(x10), s(plus(0, x10))))

  
  

  (6)    (lt(x21, x20)=true∧(∀x22:lt(x21, x20)=true ⇒ IF(true, x20, s(x22), x21)≥HELP(x20, s(x22), s(plus(x21, x22)))) ⇒ IF(true, s(x20), s(x10), s(x21))≥HELP(s(x20), s(x10), s(plus(s(x21), x10))))

  
  
  We simplified constraint (5) using rules (I), (II) which results in the following new constraint:
  

  (7)    (IF(true, s(x19), s(x10), 0)≥HELP(s(x19), s(x10), s(plus(0, x10))))

  
  
  We simplified constraint (6) using rule (VI) where we applied the induction hypothesis (∀x22:lt(x21, x20)=true ⇒ IF(true, x20, s(x22), x21)≥HELP(x20, s(x22), s(plus(x21, x22)))) with σ = [x22 / x10] which results in the following new constraint:
  

  (8)    (IF(true, x20, s(x10), x21)≥HELP(x20, s(x10), s(plus(x21, x10))) ⇒ IF(true, s(x20), s(x10), s(x21))≥HELP(s(x20), s(x10), s(plus(s(x21), x10))))

  
  
  

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

• HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)
  
  • (HELP(x3, s(x4), x8)≥IF(lt(x8, x3), x3, s(x4), x8))
    
  
  
• IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))
  
  • (IF(true, s(x19), s(x10), 0)≥HELP(s(x19), s(x10), s(plus(0, x10))))
    
  • (IF(true, x20, s(x10), x21)≥HELP(x20, s(x10), s(plus(x21, x10))) ⇒ IF(true, s(x20), s(x10), s(x21))≥HELP(s(x20), s(x10), s(plus(s(x21), x10))))
    
  
  

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(HELP(x₁, x₂, x₃)) = -1 + x₁ + x₂ - x₃   
POL(IF(x₁, x₂, x₃, x₄)) = -1 - x₁ + x₂ + x₃ - x₄   
POL(c) = -1   
POL(false) = 0   
POL(lt(x₁, x₂)) = 0   
POL(plus(x₁, x₂)) = x₁   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))

The following pairs are in P_bound:

IF(true, x, s(y), c) → HELP(x, s(y), s(plus(c, y)))

The following rules are usable:

lt(x, y) → lt(s(x), s(y))
true → lt(0, s(y))
false → lt(x, 0)
x → plus(x, 0)
s(plus(x, y)) → plus(x, s(y))

(31) Obligation:

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

HELP(x, s(y), c) → IF(lt(c, x), x, s(y), c)

The TRS R consists of the following rules:

plus(x, s(y)) → s(plus(x, y))
plus(x, 0) → x
lt(x, 0) → false
lt(0, s(y)) → true
lt(s(x), s(y)) → lt(x, y)

The set Q consists of the following terms:

lt(x0, 0)
lt(0, s(x0))
lt(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