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:

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

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)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)

The signature Sigma is {cond1, cond2}

(2) Obligation:

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

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

(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:

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND1(s(x), y) → GR(s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, x, y) → COND1(p(x), y)
COND2(false, x, y) → P(x)
GR(s(x), s(y)) → GR(x, y)
NEQ(s(x), s(y)) → NEQ(x, y)

The TRS R consists of the following rules:

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

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 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

NEQ(s(x), s(y)) → NEQ(x, y)

The TRS R consists of the following rules:

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

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:

NEQ(s(x), s(y)) → NEQ(x, y)

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

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

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].

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

(11) Obligation:

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

NEQ(s(x), s(y)) → NEQ(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:

• NEQ(s(x), s(y)) → NEQ(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:

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

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:

GR(s(x), s(y)) → GR(x, y)

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

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

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].

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

(18) 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.

(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:

• GR(s(x), s(y)) → GR(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:

COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)

The TRS R consists of the following rules:

cond1(s(x), y) → cond2(gr(s(x), y), s(x), y)
cond2(true, x, y) → cond1(y, y)
cond2(false, x, y) → cond1(p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
neq(0, 0) → false
neq(0, s(x)) → true
neq(s(x), 0) → true
neq(s(x), s(y)) → neq(x, y)
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

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:

COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))
p(0)
p(s(x0))

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].

cond1(s(x0), x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
neq(0, 0)
neq(0, s(x0))
neq(s(x0), 0)
neq(s(x0), s(x1))

(25) Obligation:

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

COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(26) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(true, x, y) → COND1(y, y) we obtained the following new rules [LPAR04]:

COND2(true, s(z0), z1) → COND1(z1, z1)

(27) Obligation:

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

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, x, y) → COND1(p(x), y)
COND2(true, s(z0), z1) → COND1(z1, z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(28) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule COND2(false, x, y) → COND1(p(x), y) at position [0] we obtained the following new rules [LPAR04]:

COND2(false, 0, y1) → COND1(0, y1)
COND2(false, s(x0), y1) → COND1(x0, y1)

(29) Obligation:

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

COND2(true, x, y) → COND1(y, y)
COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, 0, y1) → COND1(0, y1)
COND2(false, s(x0), y1) → COND1(x0, y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(30) DependencyGraphProof (EQUIVALENT transformation)

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

(31) Obligation:

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

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(32) 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.

(33) Obligation:

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

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(34) 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].

p(0)
p(s(x0))

(35) Obligation:

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

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)

The TRS R consists of the following rules:

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

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.

(36) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(true, x, y) → COND1(y, y) we obtained the following new rules [LPAR04]:

COND2(true, s(z0), z1) → COND1(z1, z1)

(37) Obligation:

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

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(false, s(x0), y1) → COND1(x0, y1)
COND2(true, s(z0), z1) → COND1(z1, z1)

The TRS R consists of the following rules:

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

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.

(38) 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 COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) the following chains were created:

• We consider the chain COND2(true, x, y) → COND1(y, y), COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) which results in the following constraint:
  

  (1)    (COND1(x3, x3)=COND1(s(x4), x5) ⇒ COND1(s(x4), x5)≥COND2(gr(s(x4), x5), s(x4), x5))

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

  (2)    (COND1(s(x4), s(x4))≥COND2(gr(s(x4), s(x4)), s(x4), s(x4)))

  
  
  
• We consider the chain COND2(false, s(x0), y1) → COND1(x0, y1), COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) which results in the following constraint:
  

  (3)    (COND1(x6, x7)=COND1(s(x8), x9) ⇒ COND1(s(x8), x9)≥COND2(gr(s(x8), x9), s(x8), x9))

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

  (4)    (COND1(s(x8), x7)≥COND2(gr(s(x8), x7), s(x8), x7))

  
  
  

For Pair COND2(true, x, y) → COND1(y, y) the following chains were created:

• We consider the chain COND1(s(x), y) → COND2(gr(s(x), y), s(x), y), COND2(true, x, y) → COND1(y, y) which results in the following constraint:
  

  (5)    (COND2(gr(s(x10), x11), s(x10), x11)=COND2(true, x12, x13) ⇒ COND2(true, x12, x13)≥COND1(x13, x13))

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

  (6)    (s(x10)=x26∧gr(x26, x11)=true ⇒ COND2(true, s(x10), x11)≥COND1(x11, x11))

  
  
  We simplified constraint (6) using rule (V) (with possible (I) afterwards) using induction on gr(x26, x11)=true which results in the following new constraints:
  

  (7)    (true=true∧s(x10)=s(x27) ⇒ COND2(true, s(x10), 0)≥COND1(0, 0))

  
  

  (8)    (gr(x29, x28)=true∧s(x10)=s(x29)∧(∀x30:gr(x29, x28)=true∧s(x30)=x29 ⇒ COND2(true, s(x30), x28)≥COND1(x28, x28)) ⇒ COND2(true, s(x10), s(x28))≥COND1(s(x28), s(x28)))

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

  (9)    (COND2(true, s(x10), 0)≥COND1(0, 0))

  
  
  We simplified constraint (8) using rules (I), (II), (III), (IV) which results in the following new constraint:
  

  (10)    (gr(x29, x28)=true ⇒ COND2(true, s(x29), s(x28))≥COND1(s(x28), s(x28)))

  
  
  We simplified constraint (10) using rule (V) (with possible (I) afterwards) using induction on gr(x29, x28)=true which results in the following new constraints:
  

  (11)    (true=true ⇒ COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))

  
  

  (12)    (gr(x34, x33)=true∧(gr(x34, x33)=true ⇒ COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33))) ⇒ COND2(true, s(s(x34)), s(s(x33)))≥COND1(s(s(x33)), s(s(x33))))

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

  (13)    (COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))

  
  
  We simplified constraint (12) using rule (VI) where we applied the induction hypothesis (gr(x34, x33)=true ⇒ COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33))) with σ = [ ] which results in the following new constraint:
  

  (14)    (COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33)) ⇒ COND2(true, s(s(x34)), s(s(x33)))≥COND1(s(s(x33)), s(s(x33))))

  
  
  

For Pair COND2(false, s(x0), y1) → COND1(x0, y1) the following chains were created:

• We consider the chain COND1(s(x), y) → COND2(gr(s(x), y), s(x), y), COND2(false, s(x0), y1) → COND1(x0, y1) which results in the following constraint:
  

  (15)    (COND2(gr(s(x18), x19), s(x18), x19)=COND2(false, s(x20), x21) ⇒ COND2(false, s(x20), x21)≥COND1(x20, x21))

  
  
  We simplified constraint (15) using rules (I), (II), (III), (VII) which results in the following new constraint:
  

  (16)    (s(x18)=x36∧gr(x36, x19)=false ⇒ COND2(false, s(x18), x19)≥COND1(x18, x19))

  
  
  We simplified constraint (16) using rule (V) (with possible (I) afterwards) using induction on gr(x36, x19)=false which results in the following new constraints:
  

  (17)    (gr(x39, x38)=false∧s(x18)=s(x39)∧(∀x40:gr(x39, x38)=false∧s(x40)=x39 ⇒ COND2(false, s(x40), x38)≥COND1(x40, x38)) ⇒ COND2(false, s(x18), s(x38))≥COND1(x18, s(x38)))

  
  

  (18)    (false=false∧s(x18)=0 ⇒ COND2(false, s(x18), x41)≥COND1(x18, x41))

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

  (19)    (gr(x39, x38)=false ⇒ COND2(false, s(x39), s(x38))≥COND1(x39, s(x38)))

  
  
  We solved constraint (18) using rules (I), (II).We simplified constraint (19) using rule (V) (with possible (I) afterwards) using induction on gr(x39, x38)=false which results in the following new constraints:
  

  (20)    (gr(x44, x43)=false∧(gr(x44, x43)=false ⇒ COND2(false, s(x44), s(x43))≥COND1(x44, s(x43))) ⇒ COND2(false, s(s(x44)), s(s(x43)))≥COND1(s(x44), s(s(x43))))

  
  

  (21)    (false=false ⇒ COND2(false, s(0), s(x45))≥COND1(0, s(x45)))

  
  
  We simplified constraint (20) using rule (VI) where we applied the induction hypothesis (gr(x44, x43)=false ⇒ COND2(false, s(x44), s(x43))≥COND1(x44, s(x43))) with σ = [ ] which results in the following new constraint:
  

  (22)    (COND2(false, s(x44), s(x43))≥COND1(x44, s(x43)) ⇒ COND2(false, s(s(x44)), s(s(x43)))≥COND1(s(x44), s(s(x43))))

  
  
  We simplified constraint (21) using rules (I), (II) which results in the following new constraint:
  

  (23)    (COND2(false, s(0), s(x45))≥COND1(0, s(x45)))

  
  
  

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

• COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
  
  • (COND1(s(x4), s(x4))≥COND2(gr(s(x4), s(x4)), s(x4), s(x4)))
    
  • (COND1(s(x8), x7)≥COND2(gr(s(x8), x7), s(x8), x7))
    
  
  
• COND2(true, x, y) → COND1(y, y)
  
  • (COND2(true, s(x10), 0)≥COND1(0, 0))
    
  • (COND2(true, s(s(x32)), s(0))≥COND1(s(0), s(0)))
    
  • (COND2(true, s(x34), s(x33))≥COND1(s(x33), s(x33)) ⇒ COND2(true, s(s(x34)), s(s(x33)))≥COND1(s(s(x33)), s(s(x33))))
    
  
  
• COND2(false, s(x0), y1) → COND1(x0, y1)
  
  • (COND2(false, s(x44), s(x43))≥COND1(x44, s(x43)) ⇒ COND2(false, s(s(x44)), s(s(x43)))≥COND1(s(x44), s(s(x43))))
    
  • (COND2(false, s(0), s(x45))≥COND1(0, s(x45)))
    
  
  

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(COND1(x₁, x₂)) = 1 + x₁ + x₂   
POL(COND2(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) = 1   

The following pairs are in P_>:

COND2(false, s(x0), y1) → COND1(x0, y1)

The following pairs are in P_bound:

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)
COND2(false, s(x0), y1) → COND1(x0, y1)

The following rules are usable:

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

(39) Obligation:

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

COND1(s(x), y) → COND2(gr(s(x), y), s(x), y)
COND2(true, x, y) → COND1(y, y)

The TRS R consists of the following rules:

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

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.

(40) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) we obtained the following new rules [LPAR04]:

COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0))

(41) Obligation:

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

COND2(true, x, y) → COND1(y, y)
COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0))

The TRS R consists of the following rules:

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

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.

(42) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND1(s(x), y) → COND2(gr(s(x), y), s(x), y) we obtained the following new rules [LPAR04]:

COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0))

(43) Obligation:

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

COND2(true, x, y) → COND1(y, y)
COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0))

The TRS R consists of the following rules:

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

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.

(44) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule COND1(s(x0), s(x0)) → COND2(gr(s(x0), s(x0)), s(x0), s(x0)) at position [0] we obtained the following new rules [LPAR04]:

COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))

(45) Obligation:

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

COND2(true, x, y) → COND1(y, y)
COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))

The TRS R consists of the following rules:

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

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.

(46) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule COND2(true, x, y) → COND1(y, y) we obtained the following new rules [LPAR04]:

COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))

(47) Obligation:

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

COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))

The TRS R consists of the following rules:

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

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.

(48) 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 COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0)) the following chains were created:

• We consider the chain COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0)), COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0)) which results in the following constraint:
  

  (1)    (COND1(s(x1), s(x1))=COND1(s(x2), s(x2)) ⇒ COND1(s(x2), s(x2))≥COND2(gr(x2, x2), s(x2), s(x2)))

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

  (2)    (COND1(s(x1), s(x1))≥COND2(gr(x1, x1), s(x1), s(x1)))

  
  
  

For Pair COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0)) the following chains were created:

• We consider the chain COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0)), COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0)) which results in the following constraint:
  

  (3)    (COND2(gr(x3, x3), s(x3), s(x3))=COND2(true, s(x4), s(x4)) ⇒ COND2(true, s(x4), s(x4))≥COND1(s(x4), s(x4)))

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

  (4)    (x3=x6∧gr(x3, x6)=true ⇒ COND2(true, s(x3), s(x3))≥COND1(s(x3), s(x3)))

  
  
  We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on gr(x3, x6)=true which results in the following new constraints:
  

  (5)    (true=true∧s(x7)=0 ⇒ COND2(true, s(s(x7)), s(s(x7)))≥COND1(s(s(x7)), s(s(x7))))

  
  

  (6)    (gr(x9, x8)=true∧s(x9)=s(x8)∧(gr(x9, x8)=true∧x9=x8 ⇒ COND2(true, s(x9), s(x9))≥COND1(s(x9), s(x9))) ⇒ COND2(true, s(s(x9)), s(s(x9)))≥COND1(s(s(x9)), s(s(x9))))

  
  
  We solved constraint (5) using rules (I), (II).We simplified constraint (6) using rules (I), (II) which results in the following new constraint:
  

  (7)    (gr(x9, x8)=true∧x9=x8∧(gr(x9, x8)=true∧x9=x8 ⇒ COND2(true, s(x9), s(x9))≥COND1(s(x9), s(x9))) ⇒ COND2(true, s(s(x9)), s(s(x9)))≥COND1(s(s(x9)), s(s(x9))))

  
  
  We simplified constraint (7) using rule (VI) where we applied the induction hypothesis (gr(x9, x8)=true∧x9=x8 ⇒ COND2(true, s(x9), s(x9))≥COND1(s(x9), s(x9))) with σ = [ ] which results in the following new constraint:
  

  (8)    (COND2(true, s(x9), s(x9))≥COND1(s(x9), s(x9)) ⇒ COND2(true, s(s(x9)), s(s(x9)))≥COND1(s(s(x9)), s(s(x9))))

  
  
  

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

• COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
  
  • (COND1(s(x1), s(x1))≥COND2(gr(x1, x1), s(x1), s(x1)))
    
  
  
• COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))
  
  • (COND2(true, s(x9), s(x9))≥COND1(s(x9), s(x9)) ⇒ COND2(true, s(s(x9)), s(s(x9)))≥COND1(s(s(x9)), s(s(x9))))
    
  
  

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(COND1(x₁, x₂)) = 1 + x₁   
POL(COND2(x₁, x₂, x₃)) = -1 - x₁ + x₃   
POL(c) = -1   
POL(false) = 1   
POL(gr(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 1   

The following pairs are in P_>:

COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))

The following pairs are in P_bound:

COND1(s(x0), s(x0)) → COND2(gr(x0, x0), s(x0), s(x0))
COND2(true, s(z0), s(z0)) → COND1(s(z0), s(z0))

The following rules are usable:

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

(49) Obligation:

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

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

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.

(50) PisEmptyProof (EQUIVALENT transformation)

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

(51) TRUE