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:

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

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

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

The TRS R 2 is

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))

The signature Sigma is {f}

(2) Obligation:

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

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(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:

F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)
F(true, x, y, z) → GT(x, plus(y, z))
F(true, x, y, z) → PLUS(y, z)
F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))
PLUS(n, s(m)) → PLUS(n, m)
GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(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 3 SCCs with 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

GT(s(u), s(v)) → GT(u, v)

The TRS R consists of the following rules:

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(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:

GT(s(u), s(v)) → GT(u, v)

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

f(true, x0, x1, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(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].

f(true, x0, x1, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

(11) Obligation:

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

GT(s(u), s(v)) → GT(u, v)

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:

• GT(s(u), s(v)) → GT(u, v)
  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:

PLUS(n, s(m)) → PLUS(n, m)

The TRS R consists of the following rules:

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(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:

PLUS(n, s(m)) → PLUS(n, m)

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

f(true, x0, x1, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(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].

f(true, x0, x1, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(x0), s(x1))

(18) 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.

(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:

• PLUS(n, s(m)) → PLUS(n, m)
  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:

F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))
F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)

The TRS R consists of the following rules:

f(true, x, y, z) → f(gt(x, plus(y, z)), x, s(y), z)
f(true, x, y, z) → f(gt(x, plus(y, z)), x, y, s(z))
plus(n, 0) → n
plus(n, s(m)) → s(plus(n, m))
gt(0, v) → false
gt(s(u), 0) → true
gt(s(u), s(v)) → gt(u, v)

The set Q consists of the following terms:

f(true, x0, x1, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(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:

F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))
F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(true, x0, x1, x2)
plus(x0, 0)
plus(x0, s(x1))
gt(0, x0)
gt(s(x0), 0)
gt(s(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].

f(true, x0, x1, x2)

(25) Obligation:

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

F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))
F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(26) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z)) we obtained the following new rules [LPAR04]:

F(true, z0, z1, s(z2)) → F(gt(z0, plus(z1, s(z2))), z0, z1, s(s(z2)))
F(true, z0, s(z1), z2) → F(gt(z0, plus(s(z1), z2)), z0, s(z1), s(z2))

(27) Obligation:

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

F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)
F(true, z0, z1, s(z2)) → F(gt(z0, plus(z1, s(z2))), z0, z1, s(s(z2)))
F(true, z0, s(z1), z2) → F(gt(z0, plus(s(z1), z2)), z0, s(z1), s(z2))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

(28) 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 F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z)) the following chains were created:

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

  (1)    (F(gt(x0, plus(x1, x2)), x0, x1, s(x2))=F(true, x3, x4, x5) ⇒ F(true, x3, x4, x5)≥F(gt(x3, plus(x4, x5)), x3, x4, s(x5)))

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

  (2)    (plus(x1, x2)=x24∧gt(x0, x24)=true ⇒ F(true, x0, x1, s(x2))≥F(gt(x0, plus(x1, s(x2))), x0, x1, s(s(x2))))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on gt(x0, x24)=true which results in the following new constraints:
  

  (3)    (true=true∧plus(x1, x2)=0 ⇒ F(true, s(x26), x1, s(x2))≥F(gt(s(x26), plus(x1, s(x2))), s(x26), x1, s(s(x2))))

  
  

  (4)    (gt(x28, x27)=true∧plus(x1, x2)=s(x27)∧(∀x29,x30:gt(x28, x27)=true∧plus(x29, x30)=x27 ⇒ F(true, x28, x29, s(x30))≥F(gt(x28, plus(x29, s(x30))), x28, x29, s(s(x30)))) ⇒ F(true, s(x28), x1, s(x2))≥F(gt(s(x28), plus(x1, s(x2))), s(x28), x1, s(s(x2))))

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

  (5)    (plus(x1, x2)=0 ⇒ F(true, s(x26), x1, s(x2))≥F(gt(s(x26), plus(x1, s(x2))), s(x26), x1, s(s(x2))))

  
  
  We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on plus(x1, x2)=s(x27) which results in the following new constraints:
  

  (6)    (x34=s(x27)∧gt(x28, x27)=true∧(∀x29,x30:gt(x28, x27)=true∧plus(x29, x30)=x27 ⇒ F(true, x28, x29, s(x30))≥F(gt(x28, plus(x29, s(x30))), x28, x29, s(s(x30)))) ⇒ F(true, s(x28), x34, s(0))≥F(gt(s(x28), plus(x34, s(0))), s(x28), x34, s(s(0))))

  
  

  (7)    (s(plus(x36, x35))=s(x27)∧gt(x28, x27)=true∧(∀x29,x30:gt(x28, x27)=true∧plus(x29, x30)=x27 ⇒ F(true, x28, x29, s(x30))≥F(gt(x28, plus(x29, s(x30))), x28, x29, s(s(x30))))∧(∀x37,x38,x39,x40:plus(x36, x35)=s(x37)∧gt(x38, x37)=true∧(∀x39,x40:gt(x38, x37)=true∧plus(x39, x40)=x37 ⇒ F(true, x38, x39, s(x40))≥F(gt(x38, plus(x39, s(x40))), x38, x39, s(s(x40)))) ⇒ F(true, s(x38), x36, s(x35))≥F(gt(s(x38), plus(x36, s(x35))), s(x38), x36, s(s(x35)))) ⇒ F(true, s(x28), x36, s(s(x35)))≥F(gt(s(x28), plus(x36, s(s(x35)))), s(x28), x36, s(s(s(x35)))))

  
  
  We simplified constraint (5) using rule (V) (with possible (I) afterwards) using induction on plus(x1, x2)=0 which results in the following new constraint:
  

  (8)    (x31=0 ⇒ F(true, s(x26), x31, s(0))≥F(gt(s(x26), plus(x31, s(0))), s(x26), x31, s(s(0))))

  
  
  We simplified constraint (8) using rule (III) which results in the following new constraint:
  

  (9)    (F(true, s(x26), 0, s(0))≥F(gt(s(x26), plus(0, s(0))), s(x26), 0, s(s(0))))

  
  
  We simplified constraint (6) using rules (III), (IV) which results in the following new constraint:
  

  (10)    (gt(x28, x27)=true ⇒ F(true, s(x28), s(x27), s(0))≥F(gt(s(x28), plus(s(x27), s(0))), s(x28), s(x27), s(s(0))))

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

  (11)    (plus(x36, x35)=x27∧gt(x28, x27)=true∧(∀x29,x30:gt(x28, x27)=true∧plus(x29, x30)=x27 ⇒ F(true, x28, x29, s(x30))≥F(gt(x28, plus(x29, s(x30))), x28, x29, s(s(x30))))∧(∀x37,x38,x39,x40:plus(x36, x35)=s(x37)∧gt(x38, x37)=true∧(∀x39,x40:gt(x38, x37)=true∧plus(x39, x40)=x37 ⇒ F(true, x38, x39, s(x40))≥F(gt(x38, plus(x39, s(x40))), x38, x39, s(s(x40)))) ⇒ F(true, s(x38), x36, s(x35))≥F(gt(s(x38), plus(x36, s(x35))), s(x38), x36, s(s(x35)))) ⇒ F(true, s(x28), x36, s(s(x35)))≥F(gt(s(x28), plus(x36, s(s(x35)))), s(x28), x36, s(s(s(x35)))))

  
  
  We simplified constraint (10) using rule (V) (with possible (I) afterwards) using induction on gt(x28, x27)=true which results in the following new constraints:
  

  (12)    (true=true ⇒ F(true, s(s(x42)), s(0), s(0))≥F(gt(s(s(x42)), plus(s(0), s(0))), s(s(x42)), s(0), s(s(0))))

  
  

  (13)    (gt(x44, x43)=true∧(gt(x44, x43)=true ⇒ F(true, s(x44), s(x43), s(0))≥F(gt(s(x44), plus(s(x43), s(0))), s(x44), s(x43), s(s(0)))) ⇒ F(true, s(s(x44)), s(s(x43)), s(0))≥F(gt(s(s(x44)), plus(s(s(x43)), s(0))), s(s(x44)), s(s(x43)), s(s(0))))

  
  
  We simplified constraint (12) using rules (I), (II) which results in the following new constraint:
  

  (14)    (F(true, s(s(x42)), s(0), s(0))≥F(gt(s(s(x42)), plus(s(0), s(0))), s(s(x42)), s(0), s(s(0))))

  
  
  We simplified constraint (13) using rule (VI) where we applied the induction hypothesis (gt(x44, x43)=true ⇒ F(true, s(x44), s(x43), s(0))≥F(gt(s(x44), plus(s(x43), s(0))), s(x44), s(x43), s(s(0)))) with σ = [ ] which results in the following new constraint:
  

  (15)    (F(true, s(x44), s(x43), s(0))≥F(gt(s(x44), plus(s(x43), s(0))), s(x44), s(x43), s(s(0))) ⇒ F(true, s(s(x44)), s(s(x43)), s(0))≥F(gt(s(s(x44)), plus(s(s(x43)), s(0))), s(s(x44)), s(s(x43)), s(s(0))))

  
  
  We simplified constraint (11) using rule (VI) where we applied the induction hypothesis (∀x29,x30:gt(x28, x27)=true∧plus(x29, x30)=x27 ⇒ F(true, x28, x29, s(x30))≥F(gt(x28, plus(x29, s(x30))), x28, x29, s(s(x30)))) with σ = [x29 / x36, x30 / x35] which results in the following new constraint:
  

  (16)    (F(true, x28, x36, s(x35))≥F(gt(x28, plus(x36, s(x35))), x28, x36, s(s(x35)))∧(∀x37,x38,x39,x40:plus(x36, x35)=s(x37)∧gt(x38, x37)=true∧(∀x39,x40:gt(x38, x37)=true∧plus(x39, x40)=x37 ⇒ F(true, x38, x39, s(x40))≥F(gt(x38, plus(x39, s(x40))), x38, x39, s(s(x40)))) ⇒ F(true, s(x38), x36, s(x35))≥F(gt(s(x38), plus(x36, s(x35))), s(x38), x36, s(s(x35)))) ⇒ F(true, s(x28), x36, s(s(x35)))≥F(gt(s(x28), plus(x36, s(s(x35)))), s(x28), x36, s(s(s(x35)))))

  
  
  We simplified constraint (16) using rule (IV) which results in the following new constraint:
  

  (17)    (F(true, x28, x36, s(x35))≥F(gt(x28, plus(x36, s(x35))), x28, x36, s(s(x35))) ⇒ F(true, s(x28), x36, s(s(x35)))≥F(gt(s(x28), plus(x36, s(s(x35)))), s(x28), x36, s(s(s(x35)))))

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

  (18)    (F(gt(x6, plus(x7, x8)), x6, s(x7), x8)=F(true, x9, x10, x11) ⇒ F(true, x9, x10, x11)≥F(gt(x9, plus(x10, x11)), x9, x10, s(x11)))

  
  
  We simplified constraint (18) using rules (I), (II), (III), (VII) which results in the following new constraint:
  

  (19)    (plus(x7, x8)=x45∧gt(x6, x45)=true ⇒ F(true, x6, s(x7), x8)≥F(gt(x6, plus(s(x7), x8)), x6, s(x7), s(x8)))

  
  
  We simplified constraint (19) using rule (V) (with possible (I) afterwards) using induction on gt(x6, x45)=true which results in the following new constraints:
  

  (20)    (true=true∧plus(x7, x8)=0 ⇒ F(true, s(x47), s(x7), x8)≥F(gt(s(x47), plus(s(x7), x8)), s(x47), s(x7), s(x8)))

  
  

  (21)    (gt(x49, x48)=true∧plus(x7, x8)=s(x48)∧(∀x50,x51:gt(x49, x48)=true∧plus(x50, x51)=x48 ⇒ F(true, x49, s(x50), x51)≥F(gt(x49, plus(s(x50), x51)), x49, s(x50), s(x51))) ⇒ F(true, s(x49), s(x7), x8)≥F(gt(s(x49), plus(s(x7), x8)), s(x49), s(x7), s(x8)))

  
  
  We simplified constraint (20) using rules (I), (II) which results in the following new constraint:
  

  (22)    (plus(x7, x8)=0 ⇒ F(true, s(x47), s(x7), x8)≥F(gt(s(x47), plus(s(x7), x8)), s(x47), s(x7), s(x8)))

  
  
  We simplified constraint (21) using rule (V) (with possible (I) afterwards) using induction on plus(x7, x8)=s(x48) which results in the following new constraints:
  

  (23)    (x55=s(x48)∧gt(x49, x48)=true∧(∀x50,x51:gt(x49, x48)=true∧plus(x50, x51)=x48 ⇒ F(true, x49, s(x50), x51)≥F(gt(x49, plus(s(x50), x51)), x49, s(x50), s(x51))) ⇒ F(true, s(x49), s(x55), 0)≥F(gt(s(x49), plus(s(x55), 0)), s(x49), s(x55), s(0)))

  
  

  (24)    (s(plus(x57, x56))=s(x48)∧gt(x49, x48)=true∧(∀x50,x51:gt(x49, x48)=true∧plus(x50, x51)=x48 ⇒ F(true, x49, s(x50), x51)≥F(gt(x49, plus(s(x50), x51)), x49, s(x50), s(x51)))∧(∀x58,x59,x60,x61:plus(x57, x56)=s(x58)∧gt(x59, x58)=true∧(∀x60,x61:gt(x59, x58)=true∧plus(x60, x61)=x58 ⇒ F(true, x59, s(x60), x61)≥F(gt(x59, plus(s(x60), x61)), x59, s(x60), s(x61))) ⇒ F(true, s(x59), s(x57), x56)≥F(gt(s(x59), plus(s(x57), x56)), s(x59), s(x57), s(x56))) ⇒ F(true, s(x49), s(x57), s(x56))≥F(gt(s(x49), plus(s(x57), s(x56))), s(x49), s(x57), s(s(x56))))

  
  
  We simplified constraint (22) using rule (V) (with possible (I) afterwards) using induction on plus(x7, x8)=0 which results in the following new constraint:
  

  (25)    (x52=0 ⇒ F(true, s(x47), s(x52), 0)≥F(gt(s(x47), plus(s(x52), 0)), s(x47), s(x52), s(0)))

  
  
  We simplified constraint (25) using rule (III) which results in the following new constraint:
  

  (26)    (F(true, s(x47), s(0), 0)≥F(gt(s(x47), plus(s(0), 0)), s(x47), s(0), s(0)))

  
  
  We simplified constraint (23) using rules (III), (IV) which results in the following new constraint:
  

  (27)    (gt(x49, x48)=true ⇒ F(true, s(x49), s(s(x48)), 0)≥F(gt(s(x49), plus(s(s(x48)), 0)), s(x49), s(s(x48)), s(0)))

  
  
  We simplified constraint (24) using rules (I), (II) which results in the following new constraint:
  

  (28)    (plus(x57, x56)=x48∧gt(x49, x48)=true∧(∀x50,x51:gt(x49, x48)=true∧plus(x50, x51)=x48 ⇒ F(true, x49, s(x50), x51)≥F(gt(x49, plus(s(x50), x51)), x49, s(x50), s(x51)))∧(∀x58,x59,x60,x61:plus(x57, x56)=s(x58)∧gt(x59, x58)=true∧(∀x60,x61:gt(x59, x58)=true∧plus(x60, x61)=x58 ⇒ F(true, x59, s(x60), x61)≥F(gt(x59, plus(s(x60), x61)), x59, s(x60), s(x61))) ⇒ F(true, s(x59), s(x57), x56)≥F(gt(s(x59), plus(s(x57), x56)), s(x59), s(x57), s(x56))) ⇒ F(true, s(x49), s(x57), s(x56))≥F(gt(s(x49), plus(s(x57), s(x56))), s(x49), s(x57), s(s(x56))))

  
  
  We simplified constraint (27) using rule (V) (with possible (I) afterwards) using induction on gt(x49, x48)=true which results in the following new constraints:
  

  (29)    (true=true ⇒ F(true, s(s(x63)), s(s(0)), 0)≥F(gt(s(s(x63)), plus(s(s(0)), 0)), s(s(x63)), s(s(0)), s(0)))

  
  

  (30)    (gt(x65, x64)=true∧(gt(x65, x64)=true ⇒ F(true, s(x65), s(s(x64)), 0)≥F(gt(s(x65), plus(s(s(x64)), 0)), s(x65), s(s(x64)), s(0))) ⇒ F(true, s(s(x65)), s(s(s(x64))), 0)≥F(gt(s(s(x65)), plus(s(s(s(x64))), 0)), s(s(x65)), s(s(s(x64))), s(0)))

  
  
  We simplified constraint (29) using rules (I), (II) which results in the following new constraint:
  

  (31)    (F(true, s(s(x63)), s(s(0)), 0)≥F(gt(s(s(x63)), plus(s(s(0)), 0)), s(s(x63)), s(s(0)), s(0)))

  
  
  We simplified constraint (30) using rule (VI) where we applied the induction hypothesis (gt(x65, x64)=true ⇒ F(true, s(x65), s(s(x64)), 0)≥F(gt(s(x65), plus(s(s(x64)), 0)), s(x65), s(s(x64)), s(0))) with σ = [ ] which results in the following new constraint:
  

  (32)    (F(true, s(x65), s(s(x64)), 0)≥F(gt(s(x65), plus(s(s(x64)), 0)), s(x65), s(s(x64)), s(0)) ⇒ F(true, s(s(x65)), s(s(s(x64))), 0)≥F(gt(s(s(x65)), plus(s(s(s(x64))), 0)), s(s(x65)), s(s(s(x64))), s(0)))

  
  
  We simplified constraint (28) using rule (VI) where we applied the induction hypothesis (∀x50,x51:gt(x49, x48)=true∧plus(x50, x51)=x48 ⇒ F(true, x49, s(x50), x51)≥F(gt(x49, plus(s(x50), x51)), x49, s(x50), s(x51))) with σ = [x50 / x57, x51 / x56] which results in the following new constraint:
  

  (33)    (F(true, x49, s(x57), x56)≥F(gt(x49, plus(s(x57), x56)), x49, s(x57), s(x56))∧(∀x58,x59,x60,x61:plus(x57, x56)=s(x58)∧gt(x59, x58)=true∧(∀x60,x61:gt(x59, x58)=true∧plus(x60, x61)=x58 ⇒ F(true, x59, s(x60), x61)≥F(gt(x59, plus(s(x60), x61)), x59, s(x60), s(x61))) ⇒ F(true, s(x59), s(x57), x56)≥F(gt(s(x59), plus(s(x57), x56)), s(x59), s(x57), s(x56))) ⇒ F(true, s(x49), s(x57), s(x56))≥F(gt(s(x49), plus(s(x57), s(x56))), s(x49), s(x57), s(s(x56))))

  
  
  We simplified constraint (33) using rule (IV) which results in the following new constraint:
  

  (34)    (F(true, x49, s(x57), x56)≥F(gt(x49, plus(s(x57), x56)), x49, s(x57), s(x56)) ⇒ F(true, s(x49), s(x57), s(x56))≥F(gt(s(x49), plus(s(x57), s(x56))), s(x49), s(x57), s(s(x56))))

  
  
  

For Pair F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z) the following chains were created:

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

  (35)    (F(gt(x12, plus(x13, x14)), x12, x13, s(x14))=F(true, x15, x16, x17) ⇒ F(true, x15, x16, x17)≥F(gt(x15, plus(x16, x17)), x15, s(x16), x17))

  
  
  We simplified constraint (35) using rules (I), (II), (III), (VII) which results in the following new constraint:
  

  (36)    (plus(x13, x14)=x66∧gt(x12, x66)=true ⇒ F(true, x12, x13, s(x14))≥F(gt(x12, plus(x13, s(x14))), x12, s(x13), s(x14)))

  
  
  We simplified constraint (36) using rule (V) (with possible (I) afterwards) using induction on gt(x12, x66)=true which results in the following new constraints:
  

  (37)    (true=true∧plus(x13, x14)=0 ⇒ F(true, s(x68), x13, s(x14))≥F(gt(s(x68), plus(x13, s(x14))), s(x68), s(x13), s(x14)))

  
  

  (38)    (gt(x70, x69)=true∧plus(x13, x14)=s(x69)∧(∀x71,x72:gt(x70, x69)=true∧plus(x71, x72)=x69 ⇒ F(true, x70, x71, s(x72))≥F(gt(x70, plus(x71, s(x72))), x70, s(x71), s(x72))) ⇒ F(true, s(x70), x13, s(x14))≥F(gt(s(x70), plus(x13, s(x14))), s(x70), s(x13), s(x14)))

  
  
  We simplified constraint (37) using rules (I), (II) which results in the following new constraint:
  

  (39)    (plus(x13, x14)=0 ⇒ F(true, s(x68), x13, s(x14))≥F(gt(s(x68), plus(x13, s(x14))), s(x68), s(x13), s(x14)))

  
  
  We simplified constraint (38) using rule (V) (with possible (I) afterwards) using induction on plus(x13, x14)=s(x69) which results in the following new constraints:
  

  (40)    (x76=s(x69)∧gt(x70, x69)=true∧(∀x71,x72:gt(x70, x69)=true∧plus(x71, x72)=x69 ⇒ F(true, x70, x71, s(x72))≥F(gt(x70, plus(x71, s(x72))), x70, s(x71), s(x72))) ⇒ F(true, s(x70), x76, s(0))≥F(gt(s(x70), plus(x76, s(0))), s(x70), s(x76), s(0)))

  
  

  (41)    (s(plus(x78, x77))=s(x69)∧gt(x70, x69)=true∧(∀x71,x72:gt(x70, x69)=true∧plus(x71, x72)=x69 ⇒ F(true, x70, x71, s(x72))≥F(gt(x70, plus(x71, s(x72))), x70, s(x71), s(x72)))∧(∀x79,x80,x81,x82:plus(x78, x77)=s(x79)∧gt(x80, x79)=true∧(∀x81,x82:gt(x80, x79)=true∧plus(x81, x82)=x79 ⇒ F(true, x80, x81, s(x82))≥F(gt(x80, plus(x81, s(x82))), x80, s(x81), s(x82))) ⇒ F(true, s(x80), x78, s(x77))≥F(gt(s(x80), plus(x78, s(x77))), s(x80), s(x78), s(x77))) ⇒ F(true, s(x70), x78, s(s(x77)))≥F(gt(s(x70), plus(x78, s(s(x77)))), s(x70), s(x78), s(s(x77))))

  
  
  We simplified constraint (39) using rule (V) (with possible (I) afterwards) using induction on plus(x13, x14)=0 which results in the following new constraint:
  

  (42)    (x73=0 ⇒ F(true, s(x68), x73, s(0))≥F(gt(s(x68), plus(x73, s(0))), s(x68), s(x73), s(0)))

  
  
  We simplified constraint (42) using rule (III) which results in the following new constraint:
  

  (43)    (F(true, s(x68), 0, s(0))≥F(gt(s(x68), plus(0, s(0))), s(x68), s(0), s(0)))

  
  
  We simplified constraint (40) using rules (III), (IV) which results in the following new constraint:
  

  (44)    (gt(x70, x69)=true ⇒ F(true, s(x70), s(x69), s(0))≥F(gt(s(x70), plus(s(x69), s(0))), s(x70), s(s(x69)), s(0)))

  
  
  We simplified constraint (41) using rules (I), (II) which results in the following new constraint:
  

  (45)    (plus(x78, x77)=x69∧gt(x70, x69)=true∧(∀x71,x72:gt(x70, x69)=true∧plus(x71, x72)=x69 ⇒ F(true, x70, x71, s(x72))≥F(gt(x70, plus(x71, s(x72))), x70, s(x71), s(x72)))∧(∀x79,x80,x81,x82:plus(x78, x77)=s(x79)∧gt(x80, x79)=true∧(∀x81,x82:gt(x80, x79)=true∧plus(x81, x82)=x79 ⇒ F(true, x80, x81, s(x82))≥F(gt(x80, plus(x81, s(x82))), x80, s(x81), s(x82))) ⇒ F(true, s(x80), x78, s(x77))≥F(gt(s(x80), plus(x78, s(x77))), s(x80), s(x78), s(x77))) ⇒ F(true, s(x70), x78, s(s(x77)))≥F(gt(s(x70), plus(x78, s(s(x77)))), s(x70), s(x78), s(s(x77))))

  
  
  We simplified constraint (44) using rule (V) (with possible (I) afterwards) using induction on gt(x70, x69)=true which results in the following new constraints:
  

  (46)    (true=true ⇒ F(true, s(s(x84)), s(0), s(0))≥F(gt(s(s(x84)), plus(s(0), s(0))), s(s(x84)), s(s(0)), s(0)))

  
  

  (47)    (gt(x86, x85)=true∧(gt(x86, x85)=true ⇒ F(true, s(x86), s(x85), s(0))≥F(gt(s(x86), plus(s(x85), s(0))), s(x86), s(s(x85)), s(0))) ⇒ F(true, s(s(x86)), s(s(x85)), s(0))≥F(gt(s(s(x86)), plus(s(s(x85)), s(0))), s(s(x86)), s(s(s(x85))), s(0)))

  
  
  We simplified constraint (46) using rules (I), (II) which results in the following new constraint:
  

  (48)    (F(true, s(s(x84)), s(0), s(0))≥F(gt(s(s(x84)), plus(s(0), s(0))), s(s(x84)), s(s(0)), s(0)))

  
  
  We simplified constraint (47) using rule (VI) where we applied the induction hypothesis (gt(x86, x85)=true ⇒ F(true, s(x86), s(x85), s(0))≥F(gt(s(x86), plus(s(x85), s(0))), s(x86), s(s(x85)), s(0))) with σ = [ ] which results in the following new constraint:
  

  (49)    (F(true, s(x86), s(x85), s(0))≥F(gt(s(x86), plus(s(x85), s(0))), s(x86), s(s(x85)), s(0)) ⇒ F(true, s(s(x86)), s(s(x85)), s(0))≥F(gt(s(s(x86)), plus(s(s(x85)), s(0))), s(s(x86)), s(s(s(x85))), s(0)))

  
  
  We simplified constraint (45) using rule (VI) where we applied the induction hypothesis (∀x71,x72:gt(x70, x69)=true∧plus(x71, x72)=x69 ⇒ F(true, x70, x71, s(x72))≥F(gt(x70, plus(x71, s(x72))), x70, s(x71), s(x72))) with σ = [x71 / x78, x72 / x77] which results in the following new constraint:
  

  (50)    (F(true, x70, x78, s(x77))≥F(gt(x70, plus(x78, s(x77))), x70, s(x78), s(x77))∧(∀x79,x80,x81,x82:plus(x78, x77)=s(x79)∧gt(x80, x79)=true∧(∀x81,x82:gt(x80, x79)=true∧plus(x81, x82)=x79 ⇒ F(true, x80, x81, s(x82))≥F(gt(x80, plus(x81, s(x82))), x80, s(x81), s(x82))) ⇒ F(true, s(x80), x78, s(x77))≥F(gt(s(x80), plus(x78, s(x77))), s(x80), s(x78), s(x77))) ⇒ F(true, s(x70), x78, s(s(x77)))≥F(gt(s(x70), plus(x78, s(s(x77)))), s(x70), s(x78), s(s(x77))))

  
  
  We simplified constraint (50) using rule (IV) which results in the following new constraint:
  

  (51)    (F(true, x70, x78, s(x77))≥F(gt(x70, plus(x78, s(x77))), x70, s(x78), s(x77)) ⇒ F(true, s(x70), x78, s(s(x77)))≥F(gt(s(x70), plus(x78, s(s(x77)))), s(x70), s(x78), s(s(x77))))

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

  (52)    (F(gt(x18, plus(x19, x20)), x18, s(x19), x20)=F(true, x21, x22, x23) ⇒ F(true, x21, x22, x23)≥F(gt(x21, plus(x22, x23)), x21, s(x22), x23))

  
  
  We simplified constraint (52) using rules (I), (II), (III), (VII) which results in the following new constraint:
  

  (53)    (plus(x19, x20)=x87∧gt(x18, x87)=true ⇒ F(true, x18, s(x19), x20)≥F(gt(x18, plus(s(x19), x20)), x18, s(s(x19)), x20))

  
  
  We simplified constraint (53) using rule (V) (with possible (I) afterwards) using induction on gt(x18, x87)=true which results in the following new constraints:
  

  (54)    (true=true∧plus(x19, x20)=0 ⇒ F(true, s(x89), s(x19), x20)≥F(gt(s(x89), plus(s(x19), x20)), s(x89), s(s(x19)), x20))

  
  

  (55)    (gt(x91, x90)=true∧plus(x19, x20)=s(x90)∧(∀x92,x93:gt(x91, x90)=true∧plus(x92, x93)=x90 ⇒ F(true, x91, s(x92), x93)≥F(gt(x91, plus(s(x92), x93)), x91, s(s(x92)), x93)) ⇒ F(true, s(x91), s(x19), x20)≥F(gt(s(x91), plus(s(x19), x20)), s(x91), s(s(x19)), x20))

  
  
  We simplified constraint (54) using rules (I), (II) which results in the following new constraint:
  

  (56)    (plus(x19, x20)=0 ⇒ F(true, s(x89), s(x19), x20)≥F(gt(s(x89), plus(s(x19), x20)), s(x89), s(s(x19)), x20))

  
  
  We simplified constraint (55) using rule (V) (with possible (I) afterwards) using induction on plus(x19, x20)=s(x90) which results in the following new constraints:
  

  (57)    (x97=s(x90)∧gt(x91, x90)=true∧(∀x92,x93:gt(x91, x90)=true∧plus(x92, x93)=x90 ⇒ F(true, x91, s(x92), x93)≥F(gt(x91, plus(s(x92), x93)), x91, s(s(x92)), x93)) ⇒ F(true, s(x91), s(x97), 0)≥F(gt(s(x91), plus(s(x97), 0)), s(x91), s(s(x97)), 0))

  
  

  (58)    (s(plus(x99, x98))=s(x90)∧gt(x91, x90)=true∧(∀x92,x93:gt(x91, x90)=true∧plus(x92, x93)=x90 ⇒ F(true, x91, s(x92), x93)≥F(gt(x91, plus(s(x92), x93)), x91, s(s(x92)), x93))∧(∀x100,x101,x102,x103:plus(x99, x98)=s(x100)∧gt(x101, x100)=true∧(∀x102,x103:gt(x101, x100)=true∧plus(x102, x103)=x100 ⇒ F(true, x101, s(x102), x103)≥F(gt(x101, plus(s(x102), x103)), x101, s(s(x102)), x103)) ⇒ F(true, s(x101), s(x99), x98)≥F(gt(s(x101), plus(s(x99), x98)), s(x101), s(s(x99)), x98)) ⇒ F(true, s(x91), s(x99), s(x98))≥F(gt(s(x91), plus(s(x99), s(x98))), s(x91), s(s(x99)), s(x98)))

  
  
  We simplified constraint (56) using rule (V) (with possible (I) afterwards) using induction on plus(x19, x20)=0 which results in the following new constraint:
  

  (59)    (x94=0 ⇒ F(true, s(x89), s(x94), 0)≥F(gt(s(x89), plus(s(x94), 0)), s(x89), s(s(x94)), 0))

  
  
  We simplified constraint (59) using rule (III) which results in the following new constraint:
  

  (60)    (F(true, s(x89), s(0), 0)≥F(gt(s(x89), plus(s(0), 0)), s(x89), s(s(0)), 0))

  
  
  We simplified constraint (57) using rules (III), (IV) which results in the following new constraint:
  

  (61)    (gt(x91, x90)=true ⇒ F(true, s(x91), s(s(x90)), 0)≥F(gt(s(x91), plus(s(s(x90)), 0)), s(x91), s(s(s(x90))), 0))

  
  
  We simplified constraint (58) using rules (I), (II) which results in the following new constraint:
  

  (62)    (plus(x99, x98)=x90∧gt(x91, x90)=true∧(∀x92,x93:gt(x91, x90)=true∧plus(x92, x93)=x90 ⇒ F(true, x91, s(x92), x93)≥F(gt(x91, plus(s(x92), x93)), x91, s(s(x92)), x93))∧(∀x100,x101,x102,x103:plus(x99, x98)=s(x100)∧gt(x101, x100)=true∧(∀x102,x103:gt(x101, x100)=true∧plus(x102, x103)=x100 ⇒ F(true, x101, s(x102), x103)≥F(gt(x101, plus(s(x102), x103)), x101, s(s(x102)), x103)) ⇒ F(true, s(x101), s(x99), x98)≥F(gt(s(x101), plus(s(x99), x98)), s(x101), s(s(x99)), x98)) ⇒ F(true, s(x91), s(x99), s(x98))≥F(gt(s(x91), plus(s(x99), s(x98))), s(x91), s(s(x99)), s(x98)))

  
  
  We simplified constraint (61) using rule (V) (with possible (I) afterwards) using induction on gt(x91, x90)=true which results in the following new constraints:
  

  (63)    (true=true ⇒ F(true, s(s(x105)), s(s(0)), 0)≥F(gt(s(s(x105)), plus(s(s(0)), 0)), s(s(x105)), s(s(s(0))), 0))

  
  

  (64)    (gt(x107, x106)=true∧(gt(x107, x106)=true ⇒ F(true, s(x107), s(s(x106)), 0)≥F(gt(s(x107), plus(s(s(x106)), 0)), s(x107), s(s(s(x106))), 0)) ⇒ F(true, s(s(x107)), s(s(s(x106))), 0)≥F(gt(s(s(x107)), plus(s(s(s(x106))), 0)), s(s(x107)), s(s(s(s(x106)))), 0))

  
  
  We simplified constraint (63) using rules (I), (II) which results in the following new constraint:
  

  (65)    (F(true, s(s(x105)), s(s(0)), 0)≥F(gt(s(s(x105)), plus(s(s(0)), 0)), s(s(x105)), s(s(s(0))), 0))

  
  
  We simplified constraint (64) using rule (VI) where we applied the induction hypothesis (gt(x107, x106)=true ⇒ F(true, s(x107), s(s(x106)), 0)≥F(gt(s(x107), plus(s(s(x106)), 0)), s(x107), s(s(s(x106))), 0)) with σ = [ ] which results in the following new constraint:
  

  (66)    (F(true, s(x107), s(s(x106)), 0)≥F(gt(s(x107), plus(s(s(x106)), 0)), s(x107), s(s(s(x106))), 0) ⇒ F(true, s(s(x107)), s(s(s(x106))), 0)≥F(gt(s(s(x107)), plus(s(s(s(x106))), 0)), s(s(x107)), s(s(s(s(x106)))), 0))

  
  
  We simplified constraint (62) using rule (VI) where we applied the induction hypothesis (∀x92,x93:gt(x91, x90)=true∧plus(x92, x93)=x90 ⇒ F(true, x91, s(x92), x93)≥F(gt(x91, plus(s(x92), x93)), x91, s(s(x92)), x93)) with σ = [x92 / x99, x93 / x98] which results in the following new constraint:
  

  (67)    (F(true, x91, s(x99), x98)≥F(gt(x91, plus(s(x99), x98)), x91, s(s(x99)), x98)∧(∀x100,x101,x102,x103:plus(x99, x98)=s(x100)∧gt(x101, x100)=true∧(∀x102,x103:gt(x101, x100)=true∧plus(x102, x103)=x100 ⇒ F(true, x101, s(x102), x103)≥F(gt(x101, plus(s(x102), x103)), x101, s(s(x102)), x103)) ⇒ F(true, s(x101), s(x99), x98)≥F(gt(s(x101), plus(s(x99), x98)), s(x101), s(s(x99)), x98)) ⇒ F(true, s(x91), s(x99), s(x98))≥F(gt(s(x91), plus(s(x99), s(x98))), s(x91), s(s(x99)), s(x98)))

  
  
  We simplified constraint (67) using rule (IV) which results in the following new constraint:
  

  (68)    (F(true, x91, s(x99), x98)≥F(gt(x91, plus(s(x99), x98)), x91, s(s(x99)), x98) ⇒ F(true, s(x91), s(x99), s(x98))≥F(gt(s(x91), plus(s(x99), s(x98))), s(x91), s(s(x99)), s(x98)))

  
  
  

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

• F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))
  
  • (F(true, s(x26), 0, s(0))≥F(gt(s(x26), plus(0, s(0))), s(x26), 0, s(s(0))))
    
  • (F(true, s(s(x42)), s(0), s(0))≥F(gt(s(s(x42)), plus(s(0), s(0))), s(s(x42)), s(0), s(s(0))))
    
  • (F(true, s(x44), s(x43), s(0))≥F(gt(s(x44), plus(s(x43), s(0))), s(x44), s(x43), s(s(0))) ⇒ F(true, s(s(x44)), s(s(x43)), s(0))≥F(gt(s(s(x44)), plus(s(s(x43)), s(0))), s(s(x44)), s(s(x43)), s(s(0))))
    
  • (F(true, x28, x36, s(x35))≥F(gt(x28, plus(x36, s(x35))), x28, x36, s(s(x35))) ⇒ F(true, s(x28), x36, s(s(x35)))≥F(gt(s(x28), plus(x36, s(s(x35)))), s(x28), x36, s(s(s(x35)))))
    
  • (F(true, s(x47), s(0), 0)≥F(gt(s(x47), plus(s(0), 0)), s(x47), s(0), s(0)))
    
  • (F(true, s(s(x63)), s(s(0)), 0)≥F(gt(s(s(x63)), plus(s(s(0)), 0)), s(s(x63)), s(s(0)), s(0)))
    
  • (F(true, s(x65), s(s(x64)), 0)≥F(gt(s(x65), plus(s(s(x64)), 0)), s(x65), s(s(x64)), s(0)) ⇒ F(true, s(s(x65)), s(s(s(x64))), 0)≥F(gt(s(s(x65)), plus(s(s(s(x64))), 0)), s(s(x65)), s(s(s(x64))), s(0)))
    
  • (F(true, x49, s(x57), x56)≥F(gt(x49, plus(s(x57), x56)), x49, s(x57), s(x56)) ⇒ F(true, s(x49), s(x57), s(x56))≥F(gt(s(x49), plus(s(x57), s(x56))), s(x49), s(x57), s(s(x56))))
    
  
  
• F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)
  
  • (F(true, s(x68), 0, s(0))≥F(gt(s(x68), plus(0, s(0))), s(x68), s(0), s(0)))
    
  • (F(true, s(s(x84)), s(0), s(0))≥F(gt(s(s(x84)), plus(s(0), s(0))), s(s(x84)), s(s(0)), s(0)))
    
  • (F(true, s(x86), s(x85), s(0))≥F(gt(s(x86), plus(s(x85), s(0))), s(x86), s(s(x85)), s(0)) ⇒ F(true, s(s(x86)), s(s(x85)), s(0))≥F(gt(s(s(x86)), plus(s(s(x85)), s(0))), s(s(x86)), s(s(s(x85))), s(0)))
    
  • (F(true, x70, x78, s(x77))≥F(gt(x70, plus(x78, s(x77))), x70, s(x78), s(x77)) ⇒ F(true, s(x70), x78, s(s(x77)))≥F(gt(s(x70), plus(x78, s(s(x77)))), s(x70), s(x78), s(s(x77))))
    
  • (F(true, s(x89), s(0), 0)≥F(gt(s(x89), plus(s(0), 0)), s(x89), s(s(0)), 0))
    
  • (F(true, s(s(x105)), s(s(0)), 0)≥F(gt(s(s(x105)), plus(s(s(0)), 0)), s(s(x105)), s(s(s(0))), 0))
    
  • (F(true, s(x107), s(s(x106)), 0)≥F(gt(s(x107), plus(s(s(x106)), 0)), s(x107), s(s(s(x106))), 0) ⇒ F(true, s(s(x107)), s(s(s(x106))), 0)≥F(gt(s(s(x107)), plus(s(s(s(x106))), 0)), s(s(x107)), s(s(s(s(x106)))), 0))
    
  • (F(true, x91, s(x99), x98)≥F(gt(x91, plus(s(x99), x98)), x91, s(s(x99)), x98) ⇒ F(true, s(x91), s(x99), s(x98))≥F(gt(s(x91), plus(s(x99), s(x98))), s(x91), s(s(x99)), s(x98)))
    
  
  

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(F(x₁, x₂, x₃, x₄)) = -1 - x₁ + x₂ - x₃ - x₄   
POL(c) = -1   
POL(false) = 1   
POL(gt(x₁, x₂)) = 0   
POL(plus(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))
F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)

The following pairs are in P_bound:

F(true, x, y, z) → F(gt(x, plus(y, z)), x, y, s(z))
F(true, x, y, z) → F(gt(x, plus(y, z)), x, s(y), z)

The following rules are usable:

false → gt(0, v)
gt(u, v) → gt(s(u), s(v))
true → gt(s(u), 0)

(29) Obligation:

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

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

The set Q consists of the following terms:

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

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

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE