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:

cond(true, x, y) → cond(gr(x, y), y, x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

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

The TRS R 2 is

cond(true, x, y) → cond(gr(x, y), y, x)

The signature Sigma is {cond}

(2) Obligation:

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

cond(true, x, y) → cond(gr(x, y), y, x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(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:

COND(true, x, y) → COND(gr(x, y), y, x)
COND(true, x, y) → GR(x, y)
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x, y) → cond(gr(x, y), y, x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(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 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) Obligation:

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

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond(true, x, y) → cond(gr(x, y), y, x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(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:

GR(s(x), s(y)) → GR(x, y)

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

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(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].

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

(11) Obligation:

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

GR(s(x), s(y)) → GR(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:

• GR(s(x), s(y)) → GR(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:

COND(true, x, y) → COND(gr(x, y), y, x)

The TRS R consists of the following rules:

cond(true, x, y) → cond(gr(x, y), y, x)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(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:

COND(true, x, y) → COND(gr(x, y), y, x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond(true, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(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].

cond(true, x0, x1)

(18) Obligation:

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

COND(true, x, y) → COND(gr(x, y), y, x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

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

(19) 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) → COND(gr(x, y), y, x) the following chains were created:

• We consider the chain COND(true, x, y) → COND(gr(x, y), y, x), COND(true, x, y) → COND(gr(x, y), y, x) which results in the following constraint:
  

  (1)    (COND(gr(x0, x1), x1, x0)=COND(true, x2, x3) ⇒ COND(true, x2, x3)≥COND(gr(x2, x3), x3, x2))

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

  (2)    (gr(x0, x1)=true ⇒ COND(true, x1, x0)≥COND(gr(x1, x0), x0, x1))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gr(x0, x1)=true which results in the following new constraints:
  

  (3)    (true=true ⇒ COND(true, 0, s(x5))≥COND(gr(0, s(x5)), s(x5), 0))

  
  

  (4)    (gr(x7, x6)=true∧(gr(x7, x6)=true ⇒ COND(true, x6, x7)≥COND(gr(x6, x7), x7, x6)) ⇒ COND(true, s(x6), s(x7))≥COND(gr(s(x6), s(x7)), s(x7), s(x6)))

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

  (5)    (COND(true, 0, s(x5))≥COND(gr(0, s(x5)), s(x5), 0))

  
  
  We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (gr(x7, x6)=true ⇒ COND(true, x6, x7)≥COND(gr(x6, x7), x7, x6)) with σ = [ ] which results in the following new constraint:
  

  (6)    (COND(true, x6, x7)≥COND(gr(x6, x7), x7, x6) ⇒ COND(true, s(x6), s(x7))≥COND(gr(s(x6), s(x7)), s(x7), s(x6)))

  
  
  

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

• COND(true, x, y) → COND(gr(x, y), y, x)
  
  • (COND(true, 0, s(x5))≥COND(gr(0, s(x5)), s(x5), 0))
    
  • (COND(true, x6, x7)≥COND(gr(x6, x7), x7, x6) ⇒ COND(true, s(x6), s(x7))≥COND(gr(s(x6), s(x7)), s(x7), s(x6)))
    
  
  

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)) = -1 - x₁ - x₂ + x₃   
POL(c) = -1   
POL(false) = 0   
POL(gr(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

COND(true, x, y) → COND(gr(x, y), y, x)

The following pairs are in P_bound:

COND(true, x, y) → COND(gr(x, y), y, x)

The following rules are usable:

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

(20) Obligation:

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

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE