(0) Obligation:

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
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)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The TRS R 2 is

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)

The signature Sigma is {cond4, cond1, cond2, cond3}

(2) Obligation:

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

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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(true, x, y) → COND2(gr(x, y), x, y)
COND1(true, x, y) → GR(x, y)
COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND2(true, x, y) → GR(x, 0)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND2(false, x, y) → GR(y, 0)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(true, x, y) → GR(x, 0)
COND3(true, x, y) → P(x)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND3(false, x, y) → AND(gr(x, 0), gr(y, 0))
COND3(false, x, y) → GR(x, 0)
COND3(false, x, y) → GR(y, 0)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(true, x, y) → GR(y, 0)
COND4(true, x, y) → P(y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND4(false, x, y) → AND(gr(x, 0), gr(y, 0))
COND4(false, x, y) → GR(x, 0)
COND4(false, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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 2 SCCs with 13 less nodes.

(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:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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:

GR(s(x), s(y)) → GR(x, y)

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

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

(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:

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(x, y), x, y)
cond2(true, x, y) → cond3(gr(x, 0), x, y)
cond2(false, x, y) → cond4(gr(y, 0), x, y)
cond3(true, x, y) → cond3(gr(x, 0), p(x), y)
cond3(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
cond4(true, x, y) → cond4(gr(y, 0), x, p(y))
cond4(false, x, y) → cond1(and(gr(x, 0), gr(y, 0)), x, y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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:

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

cond1(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)
gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
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(true, x0, x1)
cond2(true, x0, x1)
cond2(false, x0, x1)
cond3(true, x0, x1)
cond3(false, x0, x1)
cond4(true, x0, x1)
cond4(false, x0, x1)

(18) Obligation:

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

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x
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))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

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 COND2(true, x, y) → COND3(gr(x, 0), x, y) the following chains were created:
  • We consider the chain COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, x, y) → COND3(gr(x, 0), x, y) which results in the following constraint:

    (1)    (COND1(and(gr(x56, 0), gr(x57, 0)), x56, x57)=COND1(true, x58, x59)∧COND2(gr(x58, x59), x58, x59)=COND2(true, x60, x61) ⇒ COND2(true, x60, x61)≥COND3(gr(x60, 0), x60, x61))



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

    (2)    (0=x896gr(x56, x896)=x8940=x897gr(x57, x897)=x895and(x894, x895)=truex56=x58x57=x59gr(x58, x59)=trueCOND2(true, x58, x59)≥COND3(gr(x58, 0), x58, x59))



    We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on and(x894, x895)=true which results in the following new constraint:

    (3)    (true=true0=x896gr(x56, x896)=true0=x897gr(x57, x897)=truex56=x58x57=x59gr(x58, x59)=trueCOND2(true, x58, x59)≥COND3(gr(x58, 0), x58, x59))



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

    (4)    (0=x896gr(x56, x896)=true0=x897gr(x57, x897)=truex56=x58x57=x59gr(x58, x59)=trueCOND2(true, x58, x59)≥COND3(gr(x58, 0), x58, x59))



    We simplified constraint (4) using rule (V) (with possible (I) afterwards) using induction on gr(x58, x59)=true which results in the following new constraints:

    (5)    (true=true0=x896gr(x56, x896)=true0=x897gr(x57, x897)=truex56=s(x901)∧x57=0COND2(true, s(x901), 0)≥COND3(gr(s(x901), 0), s(x901), 0))


    (6)    (gr(x903, x902)=true0=x896gr(x56, x896)=true0=x897gr(x57, x897)=truex56=s(x903)∧x57=s(x902)∧(∀x904,x905,x906,x907:gr(x903, x902)=true0=x904gr(x905, x904)=true0=x906gr(x907, x906)=truex905=x903x907=x902COND2(true, x903, x902)≥COND3(gr(x903, 0), x903, x902)) ⇒ COND2(true, s(x903), s(x902))≥COND3(gr(s(x903), 0), s(x903), s(x902)))



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

    (7)    (0=x896s(x901)=x908gr(x908, x896)=true0=x8970=x909gr(x909, x897)=trueCOND2(true, s(x901), 0)≥COND3(gr(s(x901), 0), s(x901), 0))



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

    (8)    (gr(x903, x902)=true0=x896s(x903)=x922gr(x922, x896)=true0=x897s(x902)=x923gr(x923, x897)=true∧(∀x904,x905,x906,x907:gr(x903, x902)=true0=x904gr(x905, x904)=true0=x906gr(x907, x906)=truex905=x903x907=x902COND2(true, x903, x902)≥COND3(gr(x903, 0), x903, x902)) ⇒ COND2(true, s(x903), s(x902))≥COND3(gr(s(x903), 0), s(x903), s(x902)))



    We simplified constraint (7) using rule (V) (with possible (I) afterwards) using induction on gr(x908, x896)=true which results in the following new constraints:

    (9)    (true=true0=0s(x901)=s(x911)∧0=x8970=x909gr(x909, x897)=trueCOND2(true, s(x901), 0)≥COND3(gr(s(x901), 0), s(x901), 0))


    (10)    (gr(x913, x912)=true0=s(x912)∧s(x901)=s(x913)∧0=x8970=x909gr(x909, x897)=true∧(∀x914,x915,x916:gr(x913, x912)=true0=x912s(x914)=x9130=x9150=x916gr(x916, x915)=trueCOND2(true, s(x914), 0)≥COND3(gr(s(x914), 0), s(x914), 0)) ⇒ COND2(true, s(x901), 0)≥COND3(gr(s(x901), 0), s(x901), 0))



    We simplified constraint (9) using rules (I), (II), (IV) which results in the following new constraint:

    (11)    (0=x8970=x909gr(x909, x897)=trueCOND2(true, s(x901), 0)≥COND3(gr(s(x901), 0), s(x901), 0))



    We solved constraint (10) using rules (I), (II).We simplified constraint (11) using rule (V) (with possible (I) afterwards) using induction on gr(x909, x897)=true which results in the following new constraints:

    (12)    (true=true0=00=s(x918) ⇒ COND2(true, s(x901), 0)≥COND3(gr(s(x901), 0), s(x901), 0))


    (13)    (gr(x920, x919)=true0=s(x919)∧0=s(x920)∧(∀x921:gr(x920, x919)=true0=x9190=x920COND2(true, s(x921), 0)≥COND3(gr(s(x921), 0), s(x921), 0)) ⇒ COND2(true, s(x901), 0)≥COND3(gr(s(x901), 0), s(x901), 0))



    We solved constraint (12) using rules (I), (II).We solved constraint (13) using rules (I), (II).We simplified constraint (8) using rule (V) (with possible (I) afterwards) using induction on gr(x922, x896)=true which results in the following new constraints:

    (14)    (true=truegr(x903, x902)=true0=0s(x903)=s(x925)∧0=x897s(x902)=x923gr(x923, x897)=true∧(∀x904,x905,x906,x907:gr(x903, x902)=true0=x904gr(x905, x904)=true0=x906gr(x907, x906)=truex905=x903x907=x902COND2(true, x903, x902)≥COND3(gr(x903, 0), x903, x902)) ⇒ COND2(true, s(x903), s(x902))≥COND3(gr(s(x903), 0), s(x903), s(x902)))


    (15)    (gr(x927, x926)=truegr(x903, x902)=true0=s(x926)∧s(x903)=s(x927)∧0=x897s(x902)=x923gr(x923, x897)=true∧(∀x904,x905,x906,x907:gr(x903, x902)=true0=x904gr(x905, x904)=true0=x906gr(x907, x906)=truex905=x903x907=x902COND2(true, x903, x902)≥COND3(gr(x903, 0), x903, x902))∧(∀x928,x929,x930,x931,x932,x933,x934,x935:gr(x927, x926)=truegr(x928, x929)=true0=x926s(x928)=x9270=x930s(x929)=x931gr(x931, x930)=true∧(∀x932,x933,x934,x935:gr(x928, x929)=true0=x932gr(x933, x932)=true0=x934gr(x935, x934)=truex933=x928x935=x929COND2(true, x928, x929)≥COND3(gr(x928, 0), x928, x929)) ⇒ COND2(true, s(x928), s(x929))≥COND3(gr(s(x928), 0), s(x928), s(x929))) ⇒ COND2(true, s(x903), s(x902))≥COND3(gr(s(x903), 0), s(x903), s(x902)))



    We simplified constraint (14) using rules (I), (II), (IV) which results in the following new constraint:

    (16)    (gr(x903, x902)=true0=x897s(x902)=x923gr(x923, x897)=true∧(∀x904,x905,x906,x907:gr(x903, x902)=true0=x904gr(x905, x904)=true0=x906gr(x907, x906)=truex905=x903x907=x902COND2(true, x903, x902)≥COND3(gr(x903, 0), x903, x902)) ⇒ COND2(true, s(x903), s(x902))≥COND3(gr(s(x903), 0), s(x903), s(x902)))



    We solved constraint (15) using rules (I), (II).We simplified constraint (16) using rule (V) (with possible (I) afterwards) using induction on gr(x923, x897)=true which results in the following new constraints:

    (17)    (true=truegr(x903, x902)=true0=0s(x902)=s(x937)∧(∀x904,x905,x906,x907:gr(x903, x902)=true0=x904gr(x905, x904)=true0=x906gr(x907, x906)=truex905=x903x907=x902COND2(true, x903, x902)≥COND3(gr(x903, 0), x903, x902)) ⇒ COND2(true, s(x903), s(x902))≥COND3(gr(s(x903), 0), s(x903), s(x902)))


    (18)    (gr(x939, x938)=truegr(x903, x902)=true0=s(x938)∧s(x902)=s(x939)∧(∀x904,x905,x906,x907:gr(x903, x902)=true0=x904gr(x905, x904)=true0=x906gr(x907, x906)=truex905=x903x907=x902COND2(true, x903, x902)≥COND3(gr(x903, 0), x903, x902))∧(∀x940,x941,x942,x943,x944,x945:gr(x939, x938)=truegr(x940, x941)=true0=x938s(x941)=x939∧(∀x942,x943,x944,x945:gr(x940, x941)=true0=x942gr(x943, x942)=true0=x944gr(x945, x944)=truex943=x940x945=x941COND2(true, x940, x941)≥COND3(gr(x940, 0), x940, x941)) ⇒ COND2(true, s(x940), s(x941))≥COND3(gr(s(x940), 0), s(x940), s(x941))) ⇒ COND2(true, s(x903), s(x902))≥COND3(gr(s(x903), 0), s(x903), s(x902)))



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

    (19)    (gr(x903, x902)=trueCOND2(true, s(x903), s(x902))≥COND3(gr(s(x903), 0), s(x903), s(x902)))



    We solved constraint (18) using rules (I), (II).We simplified constraint (19) using rule (V) (with possible (I) afterwards) using induction on gr(x903, x902)=true which results in the following new constraints:

    (20)    (true=trueCOND2(true, s(s(x947)), s(0))≥COND3(gr(s(s(x947)), 0), s(s(x947)), s(0)))


    (21)    (gr(x949, x948)=true∧(gr(x949, x948)=trueCOND2(true, s(x949), s(x948))≥COND3(gr(s(x949), 0), s(x949), s(x948))) ⇒ COND2(true, s(s(x949)), s(s(x948)))≥COND3(gr(s(s(x949)), 0), s(s(x949)), s(s(x948))))



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

    (22)    (COND2(true, s(s(x947)), s(0))≥COND3(gr(s(s(x947)), 0), s(s(x947)), s(0)))



    We simplified constraint (21) using rule (VI) where we applied the induction hypothesis (gr(x949, x948)=trueCOND2(true, s(x949), s(x948))≥COND3(gr(s(x949), 0), s(x949), s(x948))) with σ = [ ] which results in the following new constraint:

    (23)    (COND2(true, s(x949), s(x948))≥COND3(gr(s(x949), 0), s(x949), s(x948)) ⇒ COND2(true, s(s(x949)), s(s(x948)))≥COND3(gr(s(s(x949)), 0), s(s(x949)), s(s(x948))))



  • We consider the chain COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, x, y) → COND3(gr(x, 0), x, y) which results in the following constraint:

    (24)    (COND1(and(gr(x68, 0), gr(x69, 0)), x68, x69)=COND1(true, x70, x71)∧COND2(gr(x70, x71), x70, x71)=COND2(true, x72, x73) ⇒ COND2(true, x72, x73)≥COND3(gr(x72, 0), x72, x73))



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

    (25)    (0=x952gr(x68, x952)=x9500=x953gr(x69, x953)=x951and(x950, x951)=truex68=x70x69=x71gr(x70, x71)=trueCOND2(true, x70, x71)≥COND3(gr(x70, 0), x70, x71))



    We simplified constraint (25) using rule (V) (with possible (I) afterwards) using induction on and(x950, x951)=true which results in the following new constraint:

    (26)    (true=true0=x952gr(x68, x952)=true0=x953gr(x69, x953)=truex68=x70x69=x71gr(x70, x71)=trueCOND2(true, x70, x71)≥COND3(gr(x70, 0), x70, x71))



    We simplified constraint (26) using rules (I), (II) which results in the following new constraint:

    (27)    (0=x952gr(x68, x952)=true0=x953gr(x69, x953)=truex68=x70x69=x71gr(x70, x71)=trueCOND2(true, x70, x71)≥COND3(gr(x70, 0), x70, x71))



    We simplified constraint (27) using rule (V) (with possible (I) afterwards) using induction on gr(x70, x71)=true which results in the following new constraints:

    (28)    (true=true0=x952gr(x68, x952)=true0=x953gr(x69, x953)=truex68=s(x957)∧x69=0COND2(true, s(x957), 0)≥COND3(gr(s(x957), 0), s(x957), 0))


    (29)    (gr(x959, x958)=true0=x952gr(x68, x952)=true0=x953gr(x69, x953)=truex68=s(x959)∧x69=s(x958)∧(∀x960,x961,x962,x963:gr(x959, x958)=true0=x960gr(x961, x960)=true0=x962gr(x963, x962)=truex961=x959x963=x958COND2(true, x959, x958)≥COND3(gr(x959, 0), x959, x958)) ⇒ COND2(true, s(x959), s(x958))≥COND3(gr(s(x959), 0), s(x959), s(x958)))



    We simplified constraint (28) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (30)    (0=x952s(x957)=x964gr(x964, x952)=true0=x9530=x965gr(x965, x953)=trueCOND2(true, s(x957), 0)≥COND3(gr(s(x957), 0), s(x957), 0))



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

    (31)    (gr(x959, x958)=true0=x952s(x959)=x978gr(x978, x952)=true0=x953s(x958)=x979gr(x979, x953)=true∧(∀x960,x961,x962,x963:gr(x959, x958)=true0=x960gr(x961, x960)=true0=x962gr(x963, x962)=truex961=x959x963=x958COND2(true, x959, x958)≥COND3(gr(x959, 0), x959, x958)) ⇒ COND2(true, s(x959), s(x958))≥COND3(gr(s(x959), 0), s(x959), s(x958)))



    We simplified constraint (30) using rule (V) (with possible (I) afterwards) using induction on gr(x964, x952)=true which results in the following new constraints:

    (32)    (true=true0=0s(x957)=s(x967)∧0=x9530=x965gr(x965, x953)=trueCOND2(true, s(x957), 0)≥COND3(gr(s(x957), 0), s(x957), 0))


    (33)    (gr(x969, x968)=true0=s(x968)∧s(x957)=s(x969)∧0=x9530=x965gr(x965, x953)=true∧(∀x970,x971,x972:gr(x969, x968)=true0=x968s(x970)=x9690=x9710=x972gr(x972, x971)=trueCOND2(true, s(x970), 0)≥COND3(gr(s(x970), 0), s(x970), 0)) ⇒ COND2(true, s(x957), 0)≥COND3(gr(s(x957), 0), s(x957), 0))



    We simplified constraint (32) using rules (I), (II), (IV) which results in the following new constraint:

    (34)    (0=x9530=x965gr(x965, x953)=trueCOND2(true, s(x957), 0)≥COND3(gr(s(x957), 0), s(x957), 0))



    We solved constraint (33) using rules (I), (II).We simplified constraint (34) using rule (V) (with possible (I) afterwards) using induction on gr(x965, x953)=true which results in the following new constraints:

    (35)    (true=true0=00=s(x974) ⇒ COND2(true, s(x957), 0)≥COND3(gr(s(x957), 0), s(x957), 0))


    (36)    (gr(x976, x975)=true0=s(x975)∧0=s(x976)∧(∀x977:gr(x976, x975)=true0=x9750=x976COND2(true, s(x977), 0)≥COND3(gr(s(x977), 0), s(x977), 0)) ⇒ COND2(true, s(x957), 0)≥COND3(gr(s(x957), 0), s(x957), 0))



    We solved constraint (35) using rules (I), (II).We solved constraint (36) using rules (I), (II).We simplified constraint (31) using rule (V) (with possible (I) afterwards) using induction on gr(x978, x952)=true which results in the following new constraints:

    (37)    (true=truegr(x959, x958)=true0=0s(x959)=s(x981)∧0=x953s(x958)=x979gr(x979, x953)=true∧(∀x960,x961,x962,x963:gr(x959, x958)=true0=x960gr(x961, x960)=true0=x962gr(x963, x962)=truex961=x959x963=x958COND2(true, x959, x958)≥COND3(gr(x959, 0), x959, x958)) ⇒ COND2(true, s(x959), s(x958))≥COND3(gr(s(x959), 0), s(x959), s(x958)))


    (38)    (gr(x983, x982)=truegr(x959, x958)=true0=s(x982)∧s(x959)=s(x983)∧0=x953s(x958)=x979gr(x979, x953)=true∧(∀x960,x961,x962,x963:gr(x959, x958)=true0=x960gr(x961, x960)=true0=x962gr(x963, x962)=truex961=x959x963=x958COND2(true, x959, x958)≥COND3(gr(x959, 0), x959, x958))∧(∀x984,x985,x986,x987,x988,x989,x990,x991:gr(x983, x982)=truegr(x984, x985)=true0=x982s(x984)=x9830=x986s(x985)=x987gr(x987, x986)=true∧(∀x988,x989,x990,x991:gr(x984, x985)=true0=x988gr(x989, x988)=true0=x990gr(x991, x990)=truex989=x984x991=x985COND2(true, x984, x985)≥COND3(gr(x984, 0), x984, x985)) ⇒ COND2(true, s(x984), s(x985))≥COND3(gr(s(x984), 0), s(x984), s(x985))) ⇒ COND2(true, s(x959), s(x958))≥COND3(gr(s(x959), 0), s(x959), s(x958)))



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

    (39)    (gr(x959, x958)=true0=x953s(x958)=x979gr(x979, x953)=true∧(∀x960,x961,x962,x963:gr(x959, x958)=true0=x960gr(x961, x960)=true0=x962gr(x963, x962)=truex961=x959x963=x958COND2(true, x959, x958)≥COND3(gr(x959, 0), x959, x958)) ⇒ COND2(true, s(x959), s(x958))≥COND3(gr(s(x959), 0), s(x959), s(x958)))



    We solved constraint (38) using rules (I), (II).We simplified constraint (39) using rule (V) (with possible (I) afterwards) using induction on gr(x979, x953)=true which results in the following new constraints:

    (40)    (true=truegr(x959, x958)=true0=0s(x958)=s(x993)∧(∀x960,x961,x962,x963:gr(x959, x958)=true0=x960gr(x961, x960)=true0=x962gr(x963, x962)=truex961=x959x963=x958COND2(true, x959, x958)≥COND3(gr(x959, 0), x959, x958)) ⇒ COND2(true, s(x959), s(x958))≥COND3(gr(s(x959), 0), s(x959), s(x958)))


    (41)    (gr(x995, x994)=truegr(x959, x958)=true0=s(x994)∧s(x958)=s(x995)∧(∀x960,x961,x962,x963:gr(x959, x958)=true0=x960gr(x961, x960)=true0=x962gr(x963, x962)=truex961=x959x963=x958COND2(true, x959, x958)≥COND3(gr(x959, 0), x959, x958))∧(∀x996,x997,x998,x999,x1000,x1001:gr(x995, x994)=truegr(x996, x997)=true0=x994s(x997)=x995∧(∀x998,x999,x1000,x1001:gr(x996, x997)=true0=x998gr(x999, x998)=true0=x1000gr(x1001, x1000)=truex999=x996x1001=x997COND2(true, x996, x997)≥COND3(gr(x996, 0), x996, x997)) ⇒ COND2(true, s(x996), s(x997))≥COND3(gr(s(x996), 0), s(x996), s(x997))) ⇒ COND2(true, s(x959), s(x958))≥COND3(gr(s(x959), 0), s(x959), s(x958)))



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

    (42)    (gr(x959, x958)=trueCOND2(true, s(x959), s(x958))≥COND3(gr(s(x959), 0), s(x959), s(x958)))



    We solved constraint (41) using rules (I), (II).We simplified constraint (42) using rule (V) (with possible (I) afterwards) using induction on gr(x959, x958)=true which results in the following new constraints:

    (43)    (true=trueCOND2(true, s(s(x1003)), s(0))≥COND3(gr(s(s(x1003)), 0), s(s(x1003)), s(0)))


    (44)    (gr(x1005, x1004)=true∧(gr(x1005, x1004)=trueCOND2(true, s(x1005), s(x1004))≥COND3(gr(s(x1005), 0), s(x1005), s(x1004))) ⇒ COND2(true, s(s(x1005)), s(s(x1004)))≥COND3(gr(s(s(x1005)), 0), s(s(x1005)), s(s(x1004))))



    We simplified constraint (43) using rules (I), (II) which results in the following new constraint:

    (45)    (COND2(true, s(s(x1003)), s(0))≥COND3(gr(s(s(x1003)), 0), s(s(x1003)), s(0)))



    We simplified constraint (44) using rule (VI) where we applied the induction hypothesis (gr(x1005, x1004)=trueCOND2(true, s(x1005), s(x1004))≥COND3(gr(s(x1005), 0), s(x1005), s(x1004))) with σ = [ ] which results in the following new constraint:

    (46)    (COND2(true, s(x1005), s(x1004))≥COND3(gr(s(x1005), 0), s(x1005), s(x1004)) ⇒ COND2(true, s(s(x1005)), s(s(x1004)))≥COND3(gr(s(s(x1005)), 0), s(s(x1005)), s(s(x1004))))







For Pair COND3(true, x, y) → COND3(gr(x, 0), p(x), y) the following chains were created:
  • We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, x, y) → COND3(gr(x, 0), x, y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y) which results in the following constraint:

    (47)    (COND2(gr(x132, x133), x132, x133)=COND2(true, x134, x135)∧COND3(gr(x134, 0), x134, x135)=COND3(true, x136, x137) ⇒ COND3(true, x136, x137)≥COND3(gr(x136, 0), p(x136), x137))



    We simplified constraint (47) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (48)    (gr(x132, x133)=truex132=x1340=x1006gr(x134, x1006)=trueCOND3(true, x134, x133)≥COND3(gr(x134, 0), p(x134), x133))



    We simplified constraint (48) using rule (V) (with possible (I) afterwards) using induction on gr(x132, x133)=true which results in the following new constraints:

    (49)    (true=trues(x1008)=x1340=x1006gr(x134, x1006)=trueCOND3(true, x134, 0)≥COND3(gr(x134, 0), p(x134), 0))


    (50)    (gr(x1010, x1009)=trues(x1010)=x1340=x1006gr(x134, x1006)=true∧(∀x1011,x1012:gr(x1010, x1009)=truex1010=x10110=x1012gr(x1011, x1012)=trueCOND3(true, x1011, x1009)≥COND3(gr(x1011, 0), p(x1011), x1009)) ⇒ COND3(true, x134, s(x1009))≥COND3(gr(x134, 0), p(x134), s(x1009)))



    We simplified constraint (49) using rules (I), (II) which results in the following new constraint:

    (51)    (s(x1008)=x1340=x1006gr(x134, x1006)=trueCOND3(true, x134, 0)≥COND3(gr(x134, 0), p(x134), 0))



    We simplified constraint (50) using rule (V) (with possible (I) afterwards) using induction on gr(x134, x1006)=true which results in the following new constraints:

    (52)    (true=truegr(x1010, x1009)=trues(x1010)=s(x1019)∧0=0∧(∀x1011,x1012:gr(x1010, x1009)=truex1010=x10110=x1012gr(x1011, x1012)=trueCOND3(true, x1011, x1009)≥COND3(gr(x1011, 0), p(x1011), x1009)) ⇒ COND3(true, s(x1019), s(x1009))≥COND3(gr(s(x1019), 0), p(s(x1019)), s(x1009)))


    (53)    (gr(x1021, x1020)=truegr(x1010, x1009)=trues(x1010)=s(x1021)∧0=s(x1020)∧(∀x1011,x1012:gr(x1010, x1009)=truex1010=x10110=x1012gr(x1011, x1012)=trueCOND3(true, x1011, x1009)≥COND3(gr(x1011, 0), p(x1011), x1009))∧(∀x1022,x1023,x1024,x1025:gr(x1021, x1020)=truegr(x1022, x1023)=trues(x1022)=x10210=x1020∧(∀x1024,x1025:gr(x1022, x1023)=truex1022=x10240=x1025gr(x1024, x1025)=trueCOND3(true, x1024, x1023)≥COND3(gr(x1024, 0), p(x1024), x1023)) ⇒ COND3(true, x1021, s(x1023))≥COND3(gr(x1021, 0), p(x1021), s(x1023))) ⇒ COND3(true, s(x1021), s(x1009))≥COND3(gr(s(x1021), 0), p(s(x1021)), s(x1009)))



    We simplified constraint (51) using rule (V) (with possible (I) afterwards) using induction on gr(x134, x1006)=true which results in the following new constraints:

    (54)    (true=trues(x1008)=s(x1014)∧0=0COND3(true, s(x1014), 0)≥COND3(gr(s(x1014), 0), p(s(x1014)), 0))


    (55)    (gr(x1016, x1015)=trues(x1008)=s(x1016)∧0=s(x1015)∧(∀x1017:gr(x1016, x1015)=trues(x1017)=x10160=x1015COND3(true, x1016, 0)≥COND3(gr(x1016, 0), p(x1016), 0)) ⇒ COND3(true, s(x1016), 0)≥COND3(gr(s(x1016), 0), p(s(x1016)), 0))



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

    (56)    (COND3(true, s(x1008), 0)≥COND3(gr(s(x1008), 0), p(s(x1008)), 0))



    We solved constraint (55) using rules (I), (II).We simplified constraint (52) using rules (I), (II), (III), (IV) which results in the following new constraint:

    (57)    (gr(x1010, x1009)=trueCOND3(true, s(x1010), s(x1009))≥COND3(gr(s(x1010), 0), p(s(x1010)), s(x1009)))



    We solved constraint (53) using rules (I), (II).We simplified constraint (57) using rule (V) (with possible (I) afterwards) using induction on gr(x1010, x1009)=true which results in the following new constraints:

    (58)    (true=trueCOND3(true, s(s(x1027)), s(0))≥COND3(gr(s(s(x1027)), 0), p(s(s(x1027))), s(0)))


    (59)    (gr(x1029, x1028)=true∧(gr(x1029, x1028)=trueCOND3(true, s(x1029), s(x1028))≥COND3(gr(s(x1029), 0), p(s(x1029)), s(x1028))) ⇒ COND3(true, s(s(x1029)), s(s(x1028)))≥COND3(gr(s(s(x1029)), 0), p(s(s(x1029))), s(s(x1028))))



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

    (60)    (COND3(true, s(s(x1027)), s(0))≥COND3(gr(s(s(x1027)), 0), p(s(s(x1027))), s(0)))



    We simplified constraint (59) using rule (VI) where we applied the induction hypothesis (gr(x1029, x1028)=trueCOND3(true, s(x1029), s(x1028))≥COND3(gr(s(x1029), 0), p(s(x1029)), s(x1028))) with σ = [ ] which results in the following new constraint:

    (61)    (COND3(true, s(x1029), s(x1028))≥COND3(gr(s(x1029), 0), p(s(x1029)), s(x1028)) ⇒ COND3(true, s(s(x1029)), s(s(x1028)))≥COND3(gr(s(s(x1029)), 0), p(s(s(x1029))), s(s(x1028))))



  • We consider the chain COND2(true, x, y) → COND3(gr(x, 0), x, y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y) which results in the following constraint:

    (62)    (COND3(gr(x144, 0), x144, x145)=COND3(true, x146, x147)∧COND3(gr(x146, 0), p(x146), x147)=COND3(true, x148, x149) ⇒ COND3(true, x148, x149)≥COND3(gr(x148, 0), p(x148), x149))



    We simplified constraint (62) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (63)    (0=x1030gr(x144, x1030)=truex144=x1460=x1031gr(x146, x1031)=truep(x146)=x148COND3(true, x148, x145)≥COND3(gr(x148, 0), p(x148), x145))



    We simplified constraint (63) using rule (V) (with possible (I) afterwards) using induction on gr(x144, x1030)=true which results in the following new constraints:

    (64)    (true=true0=0s(x1033)=x1460=x1031gr(x146, x1031)=truep(x146)=x148COND3(true, x148, x145)≥COND3(gr(x148, 0), p(x148), x145))


    (65)    (gr(x1035, x1034)=true0=s(x1034)∧s(x1035)=x1460=x1031gr(x146, x1031)=truep(x146)=x148∧(∀x1036,x1037,x1038,x1039:gr(x1035, x1034)=true0=x1034x1035=x10360=x1037gr(x1036, x1037)=truep(x1036)=x1038COND3(true, x1038, x1039)≥COND3(gr(x1038, 0), p(x1038), x1039)) ⇒ COND3(true, x148, x145)≥COND3(gr(x148, 0), p(x148), x145))



    We simplified constraint (64) using rules (I), (II) which results in the following new constraint:

    (66)    (s(x1033)=x1460=x1031gr(x146, x1031)=truep(x146)=x148COND3(true, x148, x145)≥COND3(gr(x148, 0), p(x148), x145))



    We solved constraint (65) using rules (I), (II).We simplified constraint (66) using rule (V) (with possible (I) afterwards) using induction on gr(x146, x1031)=true which results in the following new constraints:

    (67)    (true=trues(x1033)=s(x1041)∧0=0p(s(x1041))=x148COND3(true, x148, x145)≥COND3(gr(x148, 0), p(x148), x145))


    (68)    (gr(x1043, x1042)=trues(x1033)=s(x1043)∧0=s(x1042)∧p(s(x1043))=x148∧(∀x1044,x1045,x1046:gr(x1043, x1042)=trues(x1044)=x10430=x1042p(x1043)=x1045COND3(true, x1045, x1046)≥COND3(gr(x1045, 0), p(x1045), x1046)) ⇒ COND3(true, x148, x145)≥COND3(gr(x148, 0), p(x148), x145))



    We simplified constraint (67) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint:

    (69)    (COND3(true, x148, x145)≥COND3(gr(x148, 0), p(x148), x145))



    We solved constraint (68) using rules (I), (II).
  • We consider the chain COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y) which results in the following constraint:

    (70)    (COND3(gr(x150, 0), p(x150), x151)=COND3(true, x152, x153)∧COND3(gr(x152, 0), p(x152), x153)=COND3(true, x154, x155) ⇒ COND3(true, x154, x155)≥COND3(gr(x154, 0), p(x154), x155))



    We simplified constraint (70) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (71)    (0=x1048gr(x150, x1048)=truep(x150)=x1520=x1049gr(x152, x1049)=truep(x152)=x154COND3(true, x154, x151)≥COND3(gr(x154, 0), p(x154), x151))



    We simplified constraint (71) using rule (V) (with possible (I) afterwards) using induction on gr(x150, x1048)=true which results in the following new constraints:

    (72)    (true=true0=0p(s(x1051))=x1520=x1049gr(x152, x1049)=truep(x152)=x154COND3(true, x154, x151)≥COND3(gr(x154, 0), p(x154), x151))


    (73)    (gr(x1053, x1052)=true0=s(x1052)∧p(s(x1053))=x1520=x1049gr(x152, x1049)=truep(x152)=x154∧(∀x1054,x1055,x1056,x1057:gr(x1053, x1052)=true0=x1052p(x1053)=x10540=x1055gr(x1054, x1055)=truep(x1054)=x1056COND3(true, x1056, x1057)≥COND3(gr(x1056, 0), p(x1056), x1057)) ⇒ COND3(true, x154, x151)≥COND3(gr(x154, 0), p(x154), x151))



    We simplified constraint (72) using rules (I), (II), (VII) which results in the following new constraint:

    (74)    (s(x1051)=x1058p(x1058)=x1520=x1049gr(x152, x1049)=truep(x152)=x154COND3(true, x154, x151)≥COND3(gr(x154, 0), p(x154), x151))



    We solved constraint (73) using rules (I), (II).We simplified constraint (74) using rule (V) (with possible (I) afterwards) using induction on gr(x152, x1049)=true which results in the following new constraints:

    (75)    (true=trues(x1051)=x1058p(x1058)=s(x1060)∧0=0p(s(x1060))=x154COND3(true, x154, x151)≥COND3(gr(x154, 0), p(x154), x151))


    (76)    (gr(x1062, x1061)=trues(x1051)=x1058p(x1058)=s(x1062)∧0=s(x1061)∧p(s(x1062))=x154∧(∀x1063,x1064,x1065,x1066:gr(x1062, x1061)=trues(x1063)=x1064p(x1064)=x10620=x1061p(x1062)=x1065COND3(true, x1065, x1066)≥COND3(gr(x1065, 0), p(x1065), x1066)) ⇒ COND3(true, x154, x151)≥COND3(gr(x154, 0), p(x154), x151))



    We simplified constraint (75) using rules (I), (II), (VII) which results in the following new constraint:

    (77)    (s(x1051)=x1058p(x1058)=s(x1060)∧s(x1060)=x1067p(x1067)=x154COND3(true, x154, x151)≥COND3(gr(x154, 0), p(x154), x151))



    We solved constraint (76) using rules (I), (II).We simplified constraint (77) using rule (V) (with possible (I) afterwards) using induction on p(x1058)=s(x1060) which results in the following new constraint:

    (78)    (x1068=s(x1060)∧s(x1051)=s(x1068)∧s(x1060)=x1067p(x1067)=x154COND3(true, x154, x151)≥COND3(gr(x154, 0), p(x154), x151))



    We simplified constraint (78) using rules (I), (II), (III), (IV) which results in the following new constraint:

    (79)    (COND3(true, x154, x151)≥COND3(gr(x154, 0), p(x154), x151))







For Pair COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) the following chains were created:
  • We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(true, x, y) → COND3(gr(x, 0), x, y), COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:

    (80)    (COND2(gr(x260, x261), x260, x261)=COND2(true, x262, x263)∧COND3(gr(x262, 0), x262, x263)=COND3(false, x264, x265) ⇒ COND3(false, x264, x265)≥COND1(and(gr(x264, 0), gr(x265, 0)), x264, x265))



    We simplified constraint (80) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (81)    (gr(x260, x261)=truex260=x2620=x1069gr(x262, x1069)=falseCOND3(false, x262, x261)≥COND1(and(gr(x262, 0), gr(x261, 0)), x262, x261))



    We simplified constraint (81) using rule (V) (with possible (I) afterwards) using induction on gr(x260, x261)=true which results in the following new constraints:

    (82)    (true=trues(x1071)=x2620=x1069gr(x262, x1069)=falseCOND3(false, x262, 0)≥COND1(and(gr(x262, 0), gr(0, 0)), x262, 0))


    (83)    (gr(x1073, x1072)=trues(x1073)=x2620=x1069gr(x262, x1069)=false∧(∀x1074,x1075:gr(x1073, x1072)=truex1073=x10740=x1075gr(x1074, x1075)=falseCOND3(false, x1074, x1072)≥COND1(and(gr(x1074, 0), gr(x1072, 0)), x1074, x1072)) ⇒ COND3(false, x262, s(x1072))≥COND1(and(gr(x262, 0), gr(s(x1072), 0)), x262, s(x1072)))



    We simplified constraint (82) using rules (I), (II) which results in the following new constraint:

    (84)    (s(x1071)=x2620=x1069gr(x262, x1069)=falseCOND3(false, x262, 0)≥COND1(and(gr(x262, 0), gr(0, 0)), x262, 0))



    We simplified constraint (83) using rule (V) (with possible (I) afterwards) using induction on gr(x262, x1069)=false which results in the following new constraints:

    (85)    (false=falsegr(x1073, x1072)=trues(x1073)=00=x1081∧(∀x1074,x1075:gr(x1073, x1072)=truex1073=x10740=x1075gr(x1074, x1075)=falseCOND3(false, x1074, x1072)≥COND1(and(gr(x1074, 0), gr(x1072, 0)), x1074, x1072)) ⇒ COND3(false, 0, s(x1072))≥COND1(and(gr(0, 0), gr(s(x1072), 0)), 0, s(x1072)))


    (86)    (gr(x1084, x1083)=falsegr(x1073, x1072)=trues(x1073)=s(x1084)∧0=s(x1083)∧(∀x1074,x1075:gr(x1073, x1072)=truex1073=x10740=x1075gr(x1074, x1075)=falseCOND3(false, x1074, x1072)≥COND1(and(gr(x1074, 0), gr(x1072, 0)), x1074, x1072))∧(∀x1085,x1086,x1087,x1088:gr(x1084, x1083)=falsegr(x1085, x1086)=trues(x1085)=x10840=x1083∧(∀x1087,x1088:gr(x1085, x1086)=truex1085=x10870=x1088gr(x1087, x1088)=falseCOND3(false, x1087, x1086)≥COND1(and(gr(x1087, 0), gr(x1086, 0)), x1087, x1086)) ⇒ COND3(false, x1084, s(x1086))≥COND1(and(gr(x1084, 0), gr(s(x1086), 0)), x1084, s(x1086))) ⇒ COND3(false, s(x1084), s(x1072))≥COND1(and(gr(s(x1084), 0), gr(s(x1072), 0)), s(x1084), s(x1072)))



    We simplified constraint (84) using rule (V) (with possible (I) afterwards) using induction on gr(x262, x1069)=false which results in the following new constraints:

    (87)    (false=falses(x1071)=00=x1076COND3(false, 0, 0)≥COND1(and(gr(0, 0), gr(0, 0)), 0, 0))


    (88)    (gr(x1079, x1078)=falses(x1071)=s(x1079)∧0=s(x1078)∧(∀x1080:gr(x1079, x1078)=falses(x1080)=x10790=x1078COND3(false, x1079, 0)≥COND1(and(gr(x1079, 0), gr(0, 0)), x1079, 0)) ⇒ COND3(false, s(x1079), 0)≥COND1(and(gr(s(x1079), 0), gr(0, 0)), s(x1079), 0))



    We solved constraint (87) using rules (I), (II).We solved constraint (88) using rules (I), (II).We solved constraint (85) using rules (I), (II).We solved constraint (86) using rules (I), (II).
  • We consider the chain COND2(true, x, y) → COND3(gr(x, 0), x, y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:

    (89)    (COND3(gr(x272, 0), x272, x273)=COND3(true, x274, x275)∧COND3(gr(x274, 0), p(x274), x275)=COND3(false, x276, x277) ⇒ COND3(false, x276, x277)≥COND1(and(gr(x276, 0), gr(x277, 0)), x276, x277))



    We simplified constraint (89) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (90)    (0=x1089gr(x272, x1089)=truex272=x2740=x1090gr(x274, x1090)=falsep(x274)=x276COND3(false, x276, x273)≥COND1(and(gr(x276, 0), gr(x273, 0)), x276, x273))



    We simplified constraint (90) using rule (V) (with possible (I) afterwards) using induction on gr(x272, x1089)=true which results in the following new constraints:

    (91)    (true=true0=0s(x1092)=x2740=x1090gr(x274, x1090)=falsep(x274)=x276COND3(false, x276, x273)≥COND1(and(gr(x276, 0), gr(x273, 0)), x276, x273))


    (92)    (gr(x1094, x1093)=true0=s(x1093)∧s(x1094)=x2740=x1090gr(x274, x1090)=falsep(x274)=x276∧(∀x1095,x1096,x1097,x1098:gr(x1094, x1093)=true0=x1093x1094=x10950=x1096gr(x1095, x1096)=falsep(x1095)=x1097COND3(false, x1097, x1098)≥COND1(and(gr(x1097, 0), gr(x1098, 0)), x1097, x1098)) ⇒ COND3(false, x276, x273)≥COND1(and(gr(x276, 0), gr(x273, 0)), x276, x273))



    We simplified constraint (91) using rules (I), (II) which results in the following new constraint:

    (93)    (s(x1092)=x2740=x1090gr(x274, x1090)=falsep(x274)=x276COND3(false, x276, x273)≥COND1(and(gr(x276, 0), gr(x273, 0)), x276, x273))



    We solved constraint (92) using rules (I), (II).We simplified constraint (93) using rule (V) (with possible (I) afterwards) using induction on gr(x274, x1090)=false which results in the following new constraints:

    (94)    (false=falses(x1092)=00=x1099p(0)=x276COND3(false, x276, x273)≥COND1(and(gr(x276, 0), gr(x273, 0)), x276, x273))


    (95)    (gr(x1102, x1101)=falses(x1092)=s(x1102)∧0=s(x1101)∧p(s(x1102))=x276∧(∀x1103,x1104,x1105:gr(x1102, x1101)=falses(x1103)=x11020=x1101p(x1102)=x1104COND3(false, x1104, x1105)≥COND1(and(gr(x1104, 0), gr(x1105, 0)), x1104, x1105)) ⇒ COND3(false, x276, x273)≥COND1(and(gr(x276, 0), gr(x273, 0)), x276, x273))



    We solved constraint (94) using rules (I), (II).We solved constraint (95) using rules (I), (II).
  • We consider the chain COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:

    (96)    (COND3(gr(x278, 0), p(x278), x279)=COND3(true, x280, x281)∧COND3(gr(x280, 0), p(x280), x281)=COND3(false, x282, x283) ⇒ COND3(false, x282, x283)≥COND1(and(gr(x282, 0), gr(x283, 0)), x282, x283))



    We simplified constraint (96) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (97)    (0=x1106gr(x278, x1106)=truep(x278)=x2800=x1107gr(x280, x1107)=falsep(x280)=x282COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))



    We simplified constraint (97) using rule (V) (with possible (I) afterwards) using induction on gr(x278, x1106)=true which results in the following new constraints:

    (98)    (true=true0=0p(s(x1109))=x2800=x1107gr(x280, x1107)=falsep(x280)=x282COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))


    (99)    (gr(x1111, x1110)=true0=s(x1110)∧p(s(x1111))=x2800=x1107gr(x280, x1107)=falsep(x280)=x282∧(∀x1112,x1113,x1114,x1115:gr(x1111, x1110)=true0=x1110p(x1111)=x11120=x1113gr(x1112, x1113)=falsep(x1112)=x1114COND3(false, x1114, x1115)≥COND1(and(gr(x1114, 0), gr(x1115, 0)), x1114, x1115)) ⇒ COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))



    We simplified constraint (98) using rules (I), (II), (VII) which results in the following new constraint:

    (100)    (s(x1109)=x1116p(x1116)=x2800=x1107gr(x280, x1107)=falsep(x280)=x282COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))



    We solved constraint (99) using rules (I), (II).We simplified constraint (100) using rule (V) (with possible (I) afterwards) using induction on gr(x280, x1107)=false which results in the following new constraints:

    (101)    (false=falses(x1109)=x1116p(x1116)=00=x1117p(0)=x282COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))


    (102)    (gr(x1120, x1119)=falses(x1109)=x1116p(x1116)=s(x1120)∧0=s(x1119)∧p(s(x1120))=x282∧(∀x1121,x1122,x1123,x1124:gr(x1120, x1119)=falses(x1121)=x1122p(x1122)=x11200=x1119p(x1120)=x1123COND3(false, x1123, x1124)≥COND1(and(gr(x1123, 0), gr(x1124, 0)), x1123, x1124)) ⇒ COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))



    We simplified constraint (101) using rules (I), (II), (IV), (VII) which results in the following new constraint:

    (103)    (s(x1109)=x1116p(x1116)=00=x1125p(x1125)=x282COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))



    We solved constraint (102) using rules (I), (II).We simplified constraint (103) using rule (V) (with possible (I) afterwards) using induction on p(x1116)=0 which results in the following new constraints:

    (104)    (0=0s(x1109)=00=x1125p(x1125)=x282COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))


    (105)    (x1126=0s(x1109)=s(x1126)∧0=x1125p(x1125)=x282COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))



    We solved constraint (104) using rules (I), (II).We simplified constraint (105) using rules (I), (II), (III), (IV) which results in the following new constraint:

    (106)    (COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))







For Pair COND1(true, x, y) → COND2(gr(x, y), x, y) the following chains were created:
  • We consider the chain COND2(true, x, y) → COND3(gr(x, 0), x, y), COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:

    (107)    (COND3(gr(x416, 0), x416, x417)=COND3(false, x418, x419)∧COND1(and(gr(x418, 0), gr(x419, 0)), x418, x419)=COND1(true, x420, x421) ⇒ COND1(true, x420, x421)≥COND2(gr(x420, x421), x420, x421))



    We simplified constraint (107) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (108)    (0=x1127gr(x416, x1127)=falsex416=x4180=x1130gr(x418, x1130)=x11280=x1131gr(x419, x1131)=x1129and(x1128, x1129)=trueCOND1(true, x418, x419)≥COND2(gr(x418, x419), x418, x419))



    We simplified constraint (108) using rule (V) (with possible (I) afterwards) using induction on and(x1128, x1129)=true which results in the following new constraint:

    (109)    (true=true0=x1127gr(x416, x1127)=falsex416=x4180=x1130gr(x418, x1130)=true0=x1131gr(x419, x1131)=trueCOND1(true, x418, x419)≥COND2(gr(x418, x419), x418, x419))



    We simplified constraint (109) using rules (I), (II) which results in the following new constraint:

    (110)    (0=x1127gr(x416, x1127)=falsex416=x4180=x1130gr(x418, x1130)=true0=x1131gr(x419, x1131)=trueCOND1(true, x418, x419)≥COND2(gr(x418, x419), x418, x419))



    We simplified constraint (110) using rule (V) (with possible (I) afterwards) using induction on gr(x416, x1127)=false which results in the following new constraints:

    (111)    (false=false0=x11340=x4180=x1130gr(x418, x1130)=true0=x1131gr(x419, x1131)=trueCOND1(true, x418, x419)≥COND2(gr(x418, x419), x418, x419))


    (112)    (gr(x1137, x1136)=false0=s(x1136)∧s(x1137)=x4180=x1130gr(x418, x1130)=true0=x1131gr(x419, x1131)=true∧(∀x1138,x1139,x1140,x1141:gr(x1137, x1136)=false0=x1136x1137=x11380=x1139gr(x1138, x1139)=true0=x1140gr(x1141, x1140)=trueCOND1(true, x1138, x1141)≥COND2(gr(x1138, x1141), x1138, x1141)) ⇒ COND1(true, x418, x419)≥COND2(gr(x418, x419), x418, x419))



    We simplified constraint (111) using rules (I), (II), (IV) which results in the following new constraint:

    (113)    (0=x4180=x1130gr(x418, x1130)=true0=x1131gr(x419, x1131)=trueCOND1(true, x418, x419)≥COND2(gr(x418, x419), x418, x419))



    We solved constraint (112) using rules (I), (II).We simplified constraint (113) using rule (V) (with possible (I) afterwards) using induction on gr(x418, x1130)=true which results in the following new constraints:

    (114)    (true=true0=s(x1143)∧0=00=x1131gr(x419, x1131)=trueCOND1(true, s(x1143), x419)≥COND2(gr(s(x1143), x419), s(x1143), x419))


    (115)    (gr(x1145, x1144)=true0=s(x1145)∧0=s(x1144)∧0=x1131gr(x419, x1131)=true∧(∀x1146,x1147:gr(x1145, x1144)=true0=x11450=x11440=x1146gr(x1147, x1146)=trueCOND1(true, x1145, x1147)≥COND2(gr(x1145, x1147), x1145, x1147)) ⇒ COND1(true, s(x1145), x419)≥COND2(gr(s(x1145), x419), s(x1145), x419))



    We solved constraint (114) using rules (I), (II).We solved constraint (115) using rules (I), (II).
  • We consider the chain COND3(true, x, y) → COND3(gr(x, 0), p(x), y), COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:

    (116)    (COND3(gr(x422, 0), p(x422), x423)=COND3(false, x424, x425)∧COND1(and(gr(x424, 0), gr(x425, 0)), x424, x425)=COND1(true, x426, x427) ⇒ COND1(true, x426, x427)≥COND2(gr(x426, x427), x426, x427))



    We simplified constraint (116) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (117)    (0=x1148gr(x422, x1148)=falsep(x422)=x4240=x1151gr(x424, x1151)=x11490=x1152gr(x425, x1152)=x1150and(x1149, x1150)=trueCOND1(true, x424, x425)≥COND2(gr(x424, x425), x424, x425))



    We simplified constraint (117) using rule (V) (with possible (I) afterwards) using induction on and(x1149, x1150)=true which results in the following new constraint:

    (118)    (true=true0=x1148gr(x422, x1148)=falsep(x422)=x4240=x1151gr(x424, x1151)=true0=x1152gr(x425, x1152)=trueCOND1(true, x424, x425)≥COND2(gr(x424, x425), x424, x425))



    We simplified constraint (118) using rules (I), (II) which results in the following new constraint:

    (119)    (0=x1148gr(x422, x1148)=falsep(x422)=x4240=x1151gr(x424, x1151)=true0=x1152gr(x425, x1152)=trueCOND1(true, x424, x425)≥COND2(gr(x424, x425), x424, x425))



    We simplified constraint (119) using rule (V) (with possible (I) afterwards) using induction on gr(x422, x1148)=false which results in the following new constraints:

    (120)    (false=false0=x1155p(0)=x4240=x1151gr(x424, x1151)=true0=x1152gr(x425, x1152)=trueCOND1(true, x424, x425)≥COND2(gr(x424, x425), x424, x425))


    (121)    (gr(x1158, x1157)=false0=s(x1157)∧p(s(x1158))=x4240=x1151gr(x424, x1151)=true0=x1152gr(x425, x1152)=true∧(∀x1159,x1160,x1161,x1162:gr(x1158, x1157)=false0=x1157p(x1158)=x11590=x1160gr(x1159, x1160)=true0=x1161gr(x1162, x1161)=trueCOND1(true, x1159, x1162)≥COND2(gr(x1159, x1162), x1159, x1162)) ⇒ COND1(true, x424, x425)≥COND2(gr(x424, x425), x424, x425))



    We simplified constraint (120) using rules (I), (II), (IV), (VII) which results in the following new constraint:

    (122)    (0=x1163p(x1163)=x4240=x1151gr(x424, x1151)=true0=x1152gr(x425, x1152)=trueCOND1(true, x424, x425)≥COND2(gr(x424, x425), x424, x425))



    We solved constraint (121) using rules (I), (II).We simplified constraint (122) using rule (V) (with possible (I) afterwards) using induction on gr(x424, x1151)=true which results in the following new constraints:

    (123)    (true=true0=x1163p(x1163)=s(x1165)∧0=00=x1152gr(x425, x1152)=trueCOND1(true, s(x1165), x425)≥COND2(gr(s(x1165), x425), s(x1165), x425))


    (124)    (gr(x1167, x1166)=true0=x1163p(x1163)=s(x1167)∧0=s(x1166)∧0=x1152gr(x425, x1152)=true∧(∀x1168,x1169,x1170:gr(x1167, x1166)=true0=x1168p(x1168)=x11670=x11660=x1169gr(x1170, x1169)=trueCOND1(true, x1167, x1170)≥COND2(gr(x1167, x1170), x1167, x1170)) ⇒ COND1(true, s(x1167), x425)≥COND2(gr(s(x1167), x425), s(x1167), x425))



    We simplified constraint (123) using rules (I), (II) which results in the following new constraint:

    (125)    (0=x1163p(x1163)=s(x1165)∧0=x1152gr(x425, x1152)=trueCOND1(true, s(x1165), x425)≥COND2(gr(s(x1165), x425), s(x1165), x425))



    We solved constraint (124) using rules (I), (II).We simplified constraint (125) using rule (V) (with possible (I) afterwards) using induction on p(x1163)=s(x1165) which results in the following new constraint:

    (126)    (x1171=s(x1165)∧0=s(x1171)∧0=x1152gr(x425, x1152)=trueCOND1(true, s(x1165), x425)≥COND2(gr(s(x1165), x425), s(x1165), x425))



    We solved constraint (126) using rules (I), (II).
  • We consider the chain COND2(false, x, y) → COND4(gr(y, 0), x, y), COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:

    (127)    (COND4(gr(x499, 0), x498, x499)=COND4(false, x500, x501)∧COND1(and(gr(x500, 0), gr(x501, 0)), x500, x501)=COND1(true, x502, x503) ⇒ COND1(true, x502, x503)≥COND2(gr(x502, x503), x502, x503))



    We simplified constraint (127) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (128)    (0=x1172gr(x499, x1172)=falsex499=x5010=x1175gr(x500, x1175)=x11730=x1176gr(x501, x1176)=x1174and(x1173, x1174)=trueCOND1(true, x500, x501)≥COND2(gr(x500, x501), x500, x501))



    We simplified constraint (128) using rule (V) (with possible (I) afterwards) using induction on and(x1173, x1174)=true which results in the following new constraint:

    (129)    (true=true0=x1172gr(x499, x1172)=falsex499=x5010=x1175gr(x500, x1175)=true0=x1176gr(x501, x1176)=trueCOND1(true, x500, x501)≥COND2(gr(x500, x501), x500, x501))



    We simplified constraint (129) using rules (I), (II) which results in the following new constraint:

    (130)    (0=x1172gr(x499, x1172)=falsex499=x5010=x1175gr(x500, x1175)=true0=x1176gr(x501, x1176)=trueCOND1(true, x500, x501)≥COND2(gr(x500, x501), x500, x501))



    We simplified constraint (130) using rule (V) (with possible (I) afterwards) using induction on gr(x499, x1172)=false which results in the following new constraints:

    (131)    (false=false0=x11790=x5010=x1175gr(x500, x1175)=true0=x1176gr(x501, x1176)=trueCOND1(true, x500, x501)≥COND2(gr(x500, x501), x500, x501))


    (132)    (gr(x1182, x1181)=false0=s(x1181)∧s(x1182)=x5010=x1175gr(x500, x1175)=true0=x1176gr(x501, x1176)=true∧(∀x1183,x1184,x1185,x1186:gr(x1182, x1181)=false0=x1181x1182=x11830=x1184gr(x1185, x1184)=true0=x1186gr(x1183, x1186)=trueCOND1(true, x1185, x1183)≥COND2(gr(x1185, x1183), x1185, x1183)) ⇒ COND1(true, x500, x501)≥COND2(gr(x500, x501), x500, x501))



    We simplified constraint (131) using rules (I), (II), (IV) which results in the following new constraint:

    (133)    (0=x5010=x1175gr(x500, x1175)=true0=x1176gr(x501, x1176)=trueCOND1(true, x500, x501)≥COND2(gr(x500, x501), x500, x501))



    We solved constraint (132) using rules (I), (II).We simplified constraint (133) using rule (V) (with possible (I) afterwards) using induction on gr(x500, x1175)=true which results in the following new constraints:

    (134)    (true=true0=x5010=00=x1176gr(x501, x1176)=trueCOND1(true, s(x1188), x501)≥COND2(gr(s(x1188), x501), s(x1188), x501))


    (135)    (gr(x1190, x1189)=true0=x5010=s(x1189)∧0=x1176gr(x501, x1176)=true∧(∀x1191,x1192:gr(x1190, x1189)=true0=x11910=x11890=x1192gr(x1191, x1192)=trueCOND1(true, x1190, x1191)≥COND2(gr(x1190, x1191), x1190, x1191)) ⇒ COND1(true, s(x1190), x501)≥COND2(gr(s(x1190), x501), s(x1190), x501))



    We simplified constraint (134) using rules (I), (II) which results in the following new constraint:

    (136)    (0=x5010=x1176gr(x501, x1176)=trueCOND1(true, s(x1188), x501)≥COND2(gr(s(x1188), x501), s(x1188), x501))



    We solved constraint (135) using rules (I), (II).We simplified constraint (136) using rule (V) (with possible (I) afterwards) using induction on gr(x501, x1176)=true which results in the following new constraints:

    (137)    (true=true0=s(x1194)∧0=0COND1(true, s(x1188), s(x1194))≥COND2(gr(s(x1188), s(x1194)), s(x1188), s(x1194)))


    (138)    (gr(x1196, x1195)=true0=s(x1196)∧0=s(x1195)∧(∀x1197:gr(x1196, x1195)=true0=x11960=x1195COND1(true, s(x1197), x1196)≥COND2(gr(s(x1197), x1196), s(x1197), x1196)) ⇒ COND1(true, s(x1188), s(x1196))≥COND2(gr(s(x1188), s(x1196)), s(x1188), s(x1196)))



    We solved constraint (137) using rules (I), (II).We solved constraint (138) using rules (I), (II).
  • We consider the chain COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y) which results in the following constraint:

    (139)    (COND4(gr(x505, 0), x504, p(x505))=COND4(false, x506, x507)∧COND1(and(gr(x506, 0), gr(x507, 0)), x506, x507)=COND1(true, x508, x509) ⇒ COND1(true, x508, x509)≥COND2(gr(x508, x509), x508, x509))



    We simplified constraint (139) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (140)    (0=x1198gr(x505, x1198)=falsep(x505)=x5070=x1201gr(x506, x1201)=x11990=x1202gr(x507, x1202)=x1200and(x1199, x1200)=trueCOND1(true, x506, x507)≥COND2(gr(x506, x507), x506, x507))



    We simplified constraint (140) using rule (V) (with possible (I) afterwards) using induction on and(x1199, x1200)=true which results in the following new constraint:

    (141)    (true=true0=x1198gr(x505, x1198)=falsep(x505)=x5070=x1201gr(x506, x1201)=true0=x1202gr(x507, x1202)=trueCOND1(true, x506, x507)≥COND2(gr(x506, x507), x506, x507))



    We simplified constraint (141) using rules (I), (II) which results in the following new constraint:

    (142)    (0=x1198gr(x505, x1198)=falsep(x505)=x5070=x1201gr(x506, x1201)=true0=x1202gr(x507, x1202)=trueCOND1(true, x506, x507)≥COND2(gr(x506, x507), x506, x507))



    We simplified constraint (142) using rule (V) (with possible (I) afterwards) using induction on gr(x505, x1198)=false which results in the following new constraints:

    (143)    (false=false0=x1205p(0)=x5070=x1201gr(x506, x1201)=true0=x1202gr(x507, x1202)=trueCOND1(true, x506, x507)≥COND2(gr(x506, x507), x506, x507))


    (144)    (gr(x1208, x1207)=false0=s(x1207)∧p(s(x1208))=x5070=x1201gr(x506, x1201)=true0=x1202gr(x507, x1202)=true∧(∀x1209,x1210,x1211,x1212:gr(x1208, x1207)=false0=x1207p(x1208)=x12090=x1210gr(x1211, x1210)=true0=x1212gr(x1209, x1212)=trueCOND1(true, x1211, x1209)≥COND2(gr(x1211, x1209), x1211, x1209)) ⇒ COND1(true, x506, x507)≥COND2(gr(x506, x507), x506, x507))



    We simplified constraint (143) using rules (I), (II), (IV), (VII) which results in the following new constraint:

    (145)    (0=x1213p(x1213)=x5070=x1201gr(x506, x1201)=true0=x1202gr(x507, x1202)=trueCOND1(true, x506, x507)≥COND2(gr(x506, x507), x506, x507))



    We solved constraint (144) using rules (I), (II).We simplified constraint (145) using rule (V) (with possible (I) afterwards) using induction on gr(x506, x1201)=true which results in the following new constraints:

    (146)    (true=true0=x1213p(x1213)=x5070=00=x1202gr(x507, x1202)=trueCOND1(true, s(x1215), x507)≥COND2(gr(s(x1215), x507), s(x1215), x507))


    (147)    (gr(x1217, x1216)=true0=x1213p(x1213)=x5070=s(x1216)∧0=x1202gr(x507, x1202)=true∧(∀x1218,x1219,x1220:gr(x1217, x1216)=true0=x1218p(x1218)=x12190=x12160=x1220gr(x1219, x1220)=trueCOND1(true, x1217, x1219)≥COND2(gr(x1217, x1219), x1217, x1219)) ⇒ COND1(true, s(x1217), x507)≥COND2(gr(s(x1217), x507), s(x1217), x507))



    We simplified constraint (146) using rules (I), (II) which results in the following new constraint:

    (148)    (0=x1213p(x1213)=x5070=x1202gr(x507, x1202)=trueCOND1(true, s(x1215), x507)≥COND2(gr(s(x1215), x507), s(x1215), x507))



    We solved constraint (147) using rules (I), (II).We simplified constraint (148) using rule (V) (with possible (I) afterwards) using induction on gr(x507, x1202)=true which results in the following new constraints:

    (149)    (true=true0=x1213p(x1213)=s(x1222)∧0=0COND1(true, s(x1215), s(x1222))≥COND2(gr(s(x1215), s(x1222)), s(x1215), s(x1222)))


    (150)    (gr(x1224, x1223)=true0=x1213p(x1213)=s(x1224)∧0=s(x1223)∧(∀x1225,x1226:gr(x1224, x1223)=true0=x1225p(x1225)=x12240=x1223COND1(true, s(x1226), x1224)≥COND2(gr(s(x1226), x1224), s(x1226), x1224)) ⇒ COND1(true, s(x1215), s(x1224))≥COND2(gr(s(x1215), s(x1224)), s(x1215), s(x1224)))



    We simplified constraint (149) using rules (I), (II) which results in the following new constraint:

    (151)    (0=x1213p(x1213)=s(x1222) ⇒ COND1(true, s(x1215), s(x1222))≥COND2(gr(s(x1215), s(x1222)), s(x1215), s(x1222)))



    We solved constraint (150) using rules (I), (II).We simplified constraint (151) using rule (V) (with possible (I) afterwards) using induction on p(x1213)=s(x1222) which results in the following new constraint:

    (152)    (x1227=s(x1222)∧0=s(x1227) ⇒ COND1(true, s(x1215), s(x1222))≥COND2(gr(s(x1215), s(x1222)), s(x1215), s(x1222)))



    We solved constraint (152) using rules (I), (II).




For Pair COND2(false, x, y) → COND4(gr(y, 0), x, y) the following chains were created:
  • We consider the chain COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(false, x, y) → COND4(gr(y, 0), x, y) which results in the following constraint:

    (153)    (COND1(and(gr(x568, 0), gr(x569, 0)), x568, x569)=COND1(true, x570, x571)∧COND2(gr(x570, x571), x570, x571)=COND2(false, x572, x573) ⇒ COND2(false, x572, x573)≥COND4(gr(x573, 0), x572, x573))



    We simplified constraint (153) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (154)    (0=x1230gr(x568, x1230)=x12280=x1231gr(x569, x1231)=x1229and(x1228, x1229)=truex568=x570x569=x571gr(x570, x571)=falseCOND2(false, x570, x571)≥COND4(gr(x571, 0), x570, x571))



    We simplified constraint (154) using rule (V) (with possible (I) afterwards) using induction on and(x1228, x1229)=true which results in the following new constraint:

    (155)    (true=true0=x1230gr(x568, x1230)=true0=x1231gr(x569, x1231)=truex568=x570x569=x571gr(x570, x571)=falseCOND2(false, x570, x571)≥COND4(gr(x571, 0), x570, x571))



    We simplified constraint (155) using rules (I), (II) which results in the following new constraint:

    (156)    (0=x1230gr(x568, x1230)=true0=x1231gr(x569, x1231)=truex568=x570x569=x571gr(x570, x571)=falseCOND2(false, x570, x571)≥COND4(gr(x571, 0), x570, x571))



    We simplified constraint (156) using rule (V) (with possible (I) afterwards) using induction on gr(x570, x571)=false which results in the following new constraints:

    (157)    (false=false0=x1230gr(x568, x1230)=true0=x1231gr(x569, x1231)=truex568=0x569=x1234COND2(false, 0, x1234)≥COND4(gr(x1234, 0), 0, x1234))


    (158)    (gr(x1237, x1236)=false0=x1230gr(x568, x1230)=true0=x1231gr(x569, x1231)=truex568=s(x1237)∧x569=s(x1236)∧(∀x1238,x1239,x1240,x1241:gr(x1237, x1236)=false0=x1238gr(x1239, x1238)=true0=x1240gr(x1241, x1240)=truex1239=x1237x1241=x1236COND2(false, x1237, x1236)≥COND4(gr(x1236, 0), x1237, x1236)) ⇒ COND2(false, s(x1237), s(x1236))≥COND4(gr(s(x1236), 0), s(x1237), s(x1236)))



    We simplified constraint (157) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (159)    (0=x12300=x1242gr(x1242, x1230)=true0=x1231gr(x569, x1231)=trueCOND2(false, 0, x569)≥COND4(gr(x569, 0), 0, x569))



    We simplified constraint (158) using rules (III), (VII) which results in the following new constraint:

    (160)    (gr(x1237, x1236)=false0=x1230s(x1237)=x1249gr(x1249, x1230)=true0=x1231s(x1236)=x1250gr(x1250, x1231)=true∧(∀x1238,x1239,x1240,x1241:gr(x1237, x1236)=false0=x1238gr(x1239, x1238)=true0=x1240gr(x1241, x1240)=truex1239=x1237x1241=x1236COND2(false, x1237, x1236)≥COND4(gr(x1236, 0), x1237, x1236)) ⇒ COND2(false, s(x1237), s(x1236))≥COND4(gr(s(x1236), 0), s(x1237), s(x1236)))



    We simplified constraint (159) using rule (V) (with possible (I) afterwards) using induction on gr(x1242, x1230)=true which results in the following new constraints:

    (161)    (true=true0=00=s(x1244)∧0=x1231gr(x569, x1231)=trueCOND2(false, 0, x569)≥COND4(gr(x569, 0), 0, x569))


    (162)    (gr(x1246, x1245)=true0=s(x1245)∧0=s(x1246)∧0=x1231gr(x569, x1231)=true∧(∀x1247,x1248:gr(x1246, x1245)=true0=x12450=x12460=x1247gr(x1248, x1247)=trueCOND2(false, 0, x1248)≥COND4(gr(x1248, 0), 0, x1248)) ⇒ COND2(false, 0, x569)≥COND4(gr(x569, 0), 0, x569))



    We solved constraint (161) using rules (I), (II).We solved constraint (162) using rules (I), (II).We simplified constraint (160) using rule (V) (with possible (I) afterwards) using induction on gr(x1249, x1230)=true which results in the following new constraints:

    (163)    (true=truegr(x1237, x1236)=false0=0s(x1237)=s(x1252)∧0=x1231s(x1236)=x1250gr(x1250, x1231)=true∧(∀x1238,x1239,x1240,x1241:gr(x1237, x1236)=false0=x1238gr(x1239, x1238)=true0=x1240gr(x1241, x1240)=truex1239=x1237x1241=x1236COND2(false, x1237, x1236)≥COND4(gr(x1236, 0), x1237, x1236)) ⇒ COND2(false, s(x1237), s(x1236))≥COND4(gr(s(x1236), 0), s(x1237), s(x1236)))


    (164)    (gr(x1254, x1253)=truegr(x1237, x1236)=false0=s(x1253)∧s(x1237)=s(x1254)∧0=x1231s(x1236)=x1250gr(x1250, x1231)=true∧(∀x1238,x1239,x1240,x1241:gr(x1237, x1236)=false0=x1238gr(x1239, x1238)=true0=x1240gr(x1241, x1240)=truex1239=x1237x1241=x1236COND2(false, x1237, x1236)≥COND4(gr(x1236, 0), x1237, x1236))∧(∀x1255,x1256,x1257,x1258,x1259,x1260,x1261,x1262:gr(x1254, x1253)=truegr(x1255, x1256)=false0=x1253s(x1255)=x12540=x1257s(x1256)=x1258gr(x1258, x1257)=true∧(∀x1259,x1260,x1261,x1262:gr(x1255, x1256)=false0=x1259gr(x1260, x1259)=true0=x1261gr(x1262, x1261)=truex1260=x1255x1262=x1256COND2(false, x1255, x1256)≥COND4(gr(x1256, 0), x1255, x1256)) ⇒ COND2(false, s(x1255), s(x1256))≥COND4(gr(s(x1256), 0), s(x1255), s(x1256))) ⇒ COND2(false, s(x1237), s(x1236))≥COND4(gr(s(x1236), 0), s(x1237), s(x1236)))



    We simplified constraint (163) using rules (I), (II), (IV) which results in the following new constraint:

    (165)    (gr(x1237, x1236)=false0=x1231s(x1236)=x1250gr(x1250, x1231)=true∧(∀x1238,x1239,x1240,x1241:gr(x1237, x1236)=false0=x1238gr(x1239, x1238)=true0=x1240gr(x1241, x1240)=truex1239=x1237x1241=x1236COND2(false, x1237, x1236)≥COND4(gr(x1236, 0), x1237, x1236)) ⇒ COND2(false, s(x1237), s(x1236))≥COND4(gr(s(x1236), 0), s(x1237), s(x1236)))



    We solved constraint (164) using rules (I), (II).We simplified constraint (165) using rule (V) (with possible (I) afterwards) using induction on gr(x1250, x1231)=true which results in the following new constraints:

    (166)    (true=truegr(x1237, x1236)=false0=0s(x1236)=s(x1264)∧(∀x1238,x1239,x1240,x1241:gr(x1237, x1236)=false0=x1238gr(x1239, x1238)=true0=x1240gr(x1241, x1240)=truex1239=x1237x1241=x1236COND2(false, x1237, x1236)≥COND4(gr(x1236, 0), x1237, x1236)) ⇒ COND2(false, s(x1237), s(x1236))≥COND4(gr(s(x1236), 0), s(x1237), s(x1236)))


    (167)    (gr(x1266, x1265)=truegr(x1237, x1236)=false0=s(x1265)∧s(x1236)=s(x1266)∧(∀x1238,x1239,x1240,x1241:gr(x1237, x1236)=false0=x1238gr(x1239, x1238)=true0=x1240gr(x1241, x1240)=truex1239=x1237x1241=x1236COND2(false, x1237, x1236)≥COND4(gr(x1236, 0), x1237, x1236))∧(∀x1267,x1268,x1269,x1270,x1271,x1272:gr(x1266, x1265)=truegr(x1267, x1268)=false0=x1265s(x1268)=x1266∧(∀x1269,x1270,x1271,x1272:gr(x1267, x1268)=false0=x1269gr(x1270, x1269)=true0=x1271gr(x1272, x1271)=truex1270=x1267x1272=x1268COND2(false, x1267, x1268)≥COND4(gr(x1268, 0), x1267, x1268)) ⇒ COND2(false, s(x1267), s(x1268))≥COND4(gr(s(x1268), 0), s(x1267), s(x1268))) ⇒ COND2(false, s(x1237), s(x1236))≥COND4(gr(s(x1236), 0), s(x1237), s(x1236)))



    We simplified constraint (166) using rules (I), (II), (IV) which results in the following new constraint:

    (168)    (gr(x1237, x1236)=falseCOND2(false, s(x1237), s(x1236))≥COND4(gr(s(x1236), 0), s(x1237), s(x1236)))



    We solved constraint (167) using rules (I), (II).We simplified constraint (168) using rule (V) (with possible (I) afterwards) using induction on gr(x1237, x1236)=false which results in the following new constraints:

    (169)    (false=falseCOND2(false, s(0), s(x1273))≥COND4(gr(s(x1273), 0), s(0), s(x1273)))


    (170)    (gr(x1276, x1275)=false∧(gr(x1276, x1275)=falseCOND2(false, s(x1276), s(x1275))≥COND4(gr(s(x1275), 0), s(x1276), s(x1275))) ⇒ COND2(false, s(s(x1276)), s(s(x1275)))≥COND4(gr(s(s(x1275)), 0), s(s(x1276)), s(s(x1275))))



    We simplified constraint (169) using rules (I), (II) which results in the following new constraint:

    (171)    (COND2(false, s(0), s(x1273))≥COND4(gr(s(x1273), 0), s(0), s(x1273)))



    We simplified constraint (170) using rule (VI) where we applied the induction hypothesis (gr(x1276, x1275)=falseCOND2(false, s(x1276), s(x1275))≥COND4(gr(s(x1275), 0), s(x1276), s(x1275))) with σ = [ ] which results in the following new constraint:

    (172)    (COND2(false, s(x1276), s(x1275))≥COND4(gr(s(x1275), 0), s(x1276), s(x1275)) ⇒ COND2(false, s(s(x1276)), s(s(x1275)))≥COND4(gr(s(s(x1275)), 0), s(s(x1276)), s(s(x1275))))



  • We consider the chain COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y), COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(false, x, y) → COND4(gr(y, 0), x, y) which results in the following constraint:

    (173)    (COND1(and(gr(x580, 0), gr(x581, 0)), x580, x581)=COND1(true, x582, x583)∧COND2(gr(x582, x583), x582, x583)=COND2(false, x584, x585) ⇒ COND2(false, x584, x585)≥COND4(gr(x585, 0), x584, x585))



    We simplified constraint (173) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (174)    (0=x1279gr(x580, x1279)=x12770=x1280gr(x581, x1280)=x1278and(x1277, x1278)=truex580=x582x581=x583gr(x582, x583)=falseCOND2(false, x582, x583)≥COND4(gr(x583, 0), x582, x583))



    We simplified constraint (174) using rule (V) (with possible (I) afterwards) using induction on and(x1277, x1278)=true which results in the following new constraint:

    (175)    (true=true0=x1279gr(x580, x1279)=true0=x1280gr(x581, x1280)=truex580=x582x581=x583gr(x582, x583)=falseCOND2(false, x582, x583)≥COND4(gr(x583, 0), x582, x583))



    We simplified constraint (175) using rules (I), (II) which results in the following new constraint:

    (176)    (0=x1279gr(x580, x1279)=true0=x1280gr(x581, x1280)=truex580=x582x581=x583gr(x582, x583)=falseCOND2(false, x582, x583)≥COND4(gr(x583, 0), x582, x583))



    We simplified constraint (176) using rule (V) (with possible (I) afterwards) using induction on gr(x582, x583)=false which results in the following new constraints:

    (177)    (false=false0=x1279gr(x580, x1279)=true0=x1280gr(x581, x1280)=truex580=0x581=x1283COND2(false, 0, x1283)≥COND4(gr(x1283, 0), 0, x1283))


    (178)    (gr(x1286, x1285)=false0=x1279gr(x580, x1279)=true0=x1280gr(x581, x1280)=truex580=s(x1286)∧x581=s(x1285)∧(∀x1287,x1288,x1289,x1290:gr(x1286, x1285)=false0=x1287gr(x1288, x1287)=true0=x1289gr(x1290, x1289)=truex1288=x1286x1290=x1285COND2(false, x1286, x1285)≥COND4(gr(x1285, 0), x1286, x1285)) ⇒ COND2(false, s(x1286), s(x1285))≥COND4(gr(s(x1285), 0), s(x1286), s(x1285)))



    We simplified constraint (177) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (179)    (0=x12790=x1291gr(x1291, x1279)=true0=x1280gr(x581, x1280)=trueCOND2(false, 0, x581)≥COND4(gr(x581, 0), 0, x581))



    We simplified constraint (178) using rules (III), (VII) which results in the following new constraint:

    (180)    (gr(x1286, x1285)=false0=x1279s(x1286)=x1298gr(x1298, x1279)=true0=x1280s(x1285)=x1299gr(x1299, x1280)=true∧(∀x1287,x1288,x1289,x1290:gr(x1286, x1285)=false0=x1287gr(x1288, x1287)=true0=x1289gr(x1290, x1289)=truex1288=x1286x1290=x1285COND2(false, x1286, x1285)≥COND4(gr(x1285, 0), x1286, x1285)) ⇒ COND2(false, s(x1286), s(x1285))≥COND4(gr(s(x1285), 0), s(x1286), s(x1285)))



    We simplified constraint (179) using rule (V) (with possible (I) afterwards) using induction on gr(x1291, x1279)=true which results in the following new constraints:

    (181)    (true=true0=00=s(x1293)∧0=x1280gr(x581, x1280)=trueCOND2(false, 0, x581)≥COND4(gr(x581, 0), 0, x581))


    (182)    (gr(x1295, x1294)=true0=s(x1294)∧0=s(x1295)∧0=x1280gr(x581, x1280)=true∧(∀x1296,x1297:gr(x1295, x1294)=true0=x12940=x12950=x1296gr(x1297, x1296)=trueCOND2(false, 0, x1297)≥COND4(gr(x1297, 0), 0, x1297)) ⇒ COND2(false, 0, x581)≥COND4(gr(x581, 0), 0, x581))



    We solved constraint (181) using rules (I), (II).We solved constraint (182) using rules (I), (II).We simplified constraint (180) using rule (V) (with possible (I) afterwards) using induction on gr(x1298, x1279)=true which results in the following new constraints:

    (183)    (true=truegr(x1286, x1285)=false0=0s(x1286)=s(x1301)∧0=x1280s(x1285)=x1299gr(x1299, x1280)=true∧(∀x1287,x1288,x1289,x1290:gr(x1286, x1285)=false0=x1287gr(x1288, x1287)=true0=x1289gr(x1290, x1289)=truex1288=x1286x1290=x1285COND2(false, x1286, x1285)≥COND4(gr(x1285, 0), x1286, x1285)) ⇒ COND2(false, s(x1286), s(x1285))≥COND4(gr(s(x1285), 0), s(x1286), s(x1285)))


    (184)    (gr(x1303, x1302)=truegr(x1286, x1285)=false0=s(x1302)∧s(x1286)=s(x1303)∧0=x1280s(x1285)=x1299gr(x1299, x1280)=true∧(∀x1287,x1288,x1289,x1290:gr(x1286, x1285)=false0=x1287gr(x1288, x1287)=true0=x1289gr(x1290, x1289)=truex1288=x1286x1290=x1285COND2(false, x1286, x1285)≥COND4(gr(x1285, 0), x1286, x1285))∧(∀x1304,x1305,x1306,x1307,x1308,x1309,x1310,x1311:gr(x1303, x1302)=truegr(x1304, x1305)=false0=x1302s(x1304)=x13030=x1306s(x1305)=x1307gr(x1307, x1306)=true∧(∀x1308,x1309,x1310,x1311:gr(x1304, x1305)=false0=x1308gr(x1309, x1308)=true0=x1310gr(x1311, x1310)=truex1309=x1304x1311=x1305COND2(false, x1304, x1305)≥COND4(gr(x1305, 0), x1304, x1305)) ⇒ COND2(false, s(x1304), s(x1305))≥COND4(gr(s(x1305), 0), s(x1304), s(x1305))) ⇒ COND2(false, s(x1286), s(x1285))≥COND4(gr(s(x1285), 0), s(x1286), s(x1285)))



    We simplified constraint (183) using rules (I), (II), (IV) which results in the following new constraint:

    (185)    (gr(x1286, x1285)=false0=x1280s(x1285)=x1299gr(x1299, x1280)=true∧(∀x1287,x1288,x1289,x1290:gr(x1286, x1285)=false0=x1287gr(x1288, x1287)=true0=x1289gr(x1290, x1289)=truex1288=x1286x1290=x1285COND2(false, x1286, x1285)≥COND4(gr(x1285, 0), x1286, x1285)) ⇒ COND2(false, s(x1286), s(x1285))≥COND4(gr(s(x1285), 0), s(x1286), s(x1285)))



    We solved constraint (184) using rules (I), (II).We simplified constraint (185) using rule (V) (with possible (I) afterwards) using induction on gr(x1299, x1280)=true which results in the following new constraints:

    (186)    (true=truegr(x1286, x1285)=false0=0s(x1285)=s(x1313)∧(∀x1287,x1288,x1289,x1290:gr(x1286, x1285)=false0=x1287gr(x1288, x1287)=true0=x1289gr(x1290, x1289)=truex1288=x1286x1290=x1285COND2(false, x1286, x1285)≥COND4(gr(x1285, 0), x1286, x1285)) ⇒ COND2(false, s(x1286), s(x1285))≥COND4(gr(s(x1285), 0), s(x1286), s(x1285)))


    (187)    (gr(x1315, x1314)=truegr(x1286, x1285)=false0=s(x1314)∧s(x1285)=s(x1315)∧(∀x1287,x1288,x1289,x1290:gr(x1286, x1285)=false0=x1287gr(x1288, x1287)=true0=x1289gr(x1290, x1289)=truex1288=x1286x1290=x1285COND2(false, x1286, x1285)≥COND4(gr(x1285, 0), x1286, x1285))∧(∀x1316,x1317,x1318,x1319,x1320,x1321:gr(x1315, x1314)=truegr(x1316, x1317)=false0=x1314s(x1317)=x1315∧(∀x1318,x1319,x1320,x1321:gr(x1316, x1317)=false0=x1318gr(x1319, x1318)=true0=x1320gr(x1321, x1320)=truex1319=x1316x1321=x1317COND2(false, x1316, x1317)≥COND4(gr(x1317, 0), x1316, x1317)) ⇒ COND2(false, s(x1316), s(x1317))≥COND4(gr(s(x1317), 0), s(x1316), s(x1317))) ⇒ COND2(false, s(x1286), s(x1285))≥COND4(gr(s(x1285), 0), s(x1286), s(x1285)))



    We simplified constraint (186) using rules (I), (II), (IV) which results in the following new constraint:

    (188)    (gr(x1286, x1285)=falseCOND2(false, s(x1286), s(x1285))≥COND4(gr(s(x1285), 0), s(x1286), s(x1285)))



    We solved constraint (187) using rules (I), (II).We simplified constraint (188) using rule (V) (with possible (I) afterwards) using induction on gr(x1286, x1285)=false which results in the following new constraints:

    (189)    (false=falseCOND2(false, s(0), s(x1322))≥COND4(gr(s(x1322), 0), s(0), s(x1322)))


    (190)    (gr(x1325, x1324)=false∧(gr(x1325, x1324)=falseCOND2(false, s(x1325), s(x1324))≥COND4(gr(s(x1324), 0), s(x1325), s(x1324))) ⇒ COND2(false, s(s(x1325)), s(s(x1324)))≥COND4(gr(s(s(x1324)), 0), s(s(x1325)), s(s(x1324))))



    We simplified constraint (189) using rules (I), (II) which results in the following new constraint:

    (191)    (COND2(false, s(0), s(x1322))≥COND4(gr(s(x1322), 0), s(0), s(x1322)))



    We simplified constraint (190) using rule (VI) where we applied the induction hypothesis (gr(x1325, x1324)=falseCOND2(false, s(x1325), s(x1324))≥COND4(gr(s(x1324), 0), s(x1325), s(x1324))) with σ = [ ] which results in the following new constraint:

    (192)    (COND2(false, s(x1325), s(x1324))≥COND4(gr(s(x1324), 0), s(x1325), s(x1324)) ⇒ COND2(false, s(s(x1325)), s(s(x1324)))≥COND4(gr(s(s(x1324)), 0), s(s(x1325)), s(s(x1324))))







For Pair COND4(true, x, y) → COND4(gr(y, 0), x, p(y)) the following chains were created:
  • We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(false, x, y) → COND4(gr(y, 0), x, y), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)) which results in the following constraint:

    (193)    (COND2(gr(x714, x715), x714, x715)=COND2(false, x716, x717)∧COND4(gr(x717, 0), x716, x717)=COND4(true, x718, x719) ⇒ COND4(true, x718, x719)≥COND4(gr(x719, 0), x718, p(x719)))



    We simplified constraint (193) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (194)    (gr(x714, x715)=falsex715=x7170=x1326gr(x717, x1326)=trueCOND4(true, x714, x717)≥COND4(gr(x717, 0), x714, p(x717)))



    We simplified constraint (194) using rule (V) (with possible (I) afterwards) using induction on gr(x714, x715)=false which results in the following new constraints:

    (195)    (false=falsex1327=x7170=x1326gr(x717, x1326)=trueCOND4(true, 0, x717)≥COND4(gr(x717, 0), 0, p(x717)))


    (196)    (gr(x1330, x1329)=falses(x1329)=x7170=x1326gr(x717, x1326)=true∧(∀x1331,x1332:gr(x1330, x1329)=falsex1329=x13310=x1332gr(x1331, x1332)=trueCOND4(true, x1330, x1331)≥COND4(gr(x1331, 0), x1330, p(x1331))) ⇒ COND4(true, s(x1330), x717)≥COND4(gr(x717, 0), s(x1330), p(x717)))



    We simplified constraint (195) using rules (I), (II), (III) which results in the following new constraint:

    (197)    (0=x1326gr(x717, x1326)=trueCOND4(true, 0, x717)≥COND4(gr(x717, 0), 0, p(x717)))



    We simplified constraint (196) using rule (V) (with possible (I) afterwards) using induction on gr(x717, x1326)=true which results in the following new constraints:

    (198)    (true=truegr(x1330, x1329)=falses(x1329)=s(x1338)∧0=0∧(∀x1331,x1332:gr(x1330, x1329)=falsex1329=x13310=x1332gr(x1331, x1332)=trueCOND4(true, x1330, x1331)≥COND4(gr(x1331, 0), x1330, p(x1331))) ⇒ COND4(true, s(x1330), s(x1338))≥COND4(gr(s(x1338), 0), s(x1330), p(s(x1338))))


    (199)    (gr(x1340, x1339)=truegr(x1330, x1329)=falses(x1329)=s(x1340)∧0=s(x1339)∧(∀x1331,x1332:gr(x1330, x1329)=falsex1329=x13310=x1332gr(x1331, x1332)=trueCOND4(true, x1330, x1331)≥COND4(gr(x1331, 0), x1330, p(x1331)))∧(∀x1341,x1342,x1343,x1344:gr(x1340, x1339)=truegr(x1341, x1342)=falses(x1342)=x13400=x1339∧(∀x1343,x1344:gr(x1341, x1342)=falsex1342=x13430=x1344gr(x1343, x1344)=trueCOND4(true, x1341, x1343)≥COND4(gr(x1343, 0), x1341, p(x1343))) ⇒ COND4(true, s(x1341), x1340)≥COND4(gr(x1340, 0), s(x1341), p(x1340))) ⇒ COND4(true, s(x1330), s(x1340))≥COND4(gr(s(x1340), 0), s(x1330), p(s(x1340))))



    We simplified constraint (197) using rule (V) (with possible (I) afterwards) using induction on gr(x717, x1326)=true which results in the following new constraints:

    (200)    (true=true0=0COND4(true, 0, s(x1334))≥COND4(gr(s(x1334), 0), 0, p(s(x1334))))


    (201)    (gr(x1336, x1335)=true0=s(x1335)∧(gr(x1336, x1335)=true0=x1335COND4(true, 0, x1336)≥COND4(gr(x1336, 0), 0, p(x1336))) ⇒ COND4(true, 0, s(x1336))≥COND4(gr(s(x1336), 0), 0, p(s(x1336))))



    We simplified constraint (200) using rules (I), (II) which results in the following new constraint:

    (202)    (COND4(true, 0, s(x1334))≥COND4(gr(s(x1334), 0), 0, p(s(x1334))))



    We solved constraint (201) using rules (I), (II).We simplified constraint (198) using rules (I), (II), (III), (IV) which results in the following new constraint:

    (203)    (gr(x1330, x1329)=falseCOND4(true, s(x1330), s(x1329))≥COND4(gr(s(x1329), 0), s(x1330), p(s(x1329))))



    We solved constraint (199) using rules (I), (II).We simplified constraint (203) using rule (V) (with possible (I) afterwards) using induction on gr(x1330, x1329)=false which results in the following new constraints:

    (204)    (false=falseCOND4(true, s(0), s(x1345))≥COND4(gr(s(x1345), 0), s(0), p(s(x1345))))


    (205)    (gr(x1348, x1347)=false∧(gr(x1348, x1347)=falseCOND4(true, s(x1348), s(x1347))≥COND4(gr(s(x1347), 0), s(x1348), p(s(x1347)))) ⇒ COND4(true, s(s(x1348)), s(s(x1347)))≥COND4(gr(s(s(x1347)), 0), s(s(x1348)), p(s(s(x1347)))))



    We simplified constraint (204) using rules (I), (II) which results in the following new constraint:

    (206)    (COND4(true, s(0), s(x1345))≥COND4(gr(s(x1345), 0), s(0), p(s(x1345))))



    We simplified constraint (205) using rule (VI) where we applied the induction hypothesis (gr(x1348, x1347)=falseCOND4(true, s(x1348), s(x1347))≥COND4(gr(s(x1347), 0), s(x1348), p(s(x1347)))) with σ = [ ] which results in the following new constraint:

    (207)    (COND4(true, s(x1348), s(x1347))≥COND4(gr(s(x1347), 0), s(x1348), p(s(x1347))) ⇒ COND4(true, s(s(x1348)), s(s(x1347)))≥COND4(gr(s(s(x1347)), 0), s(s(x1348)), p(s(s(x1347)))))



  • We consider the chain COND2(false, x, y) → COND4(gr(y, 0), x, y), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)) which results in the following constraint:

    (208)    (COND4(gr(x735, 0), x734, x735)=COND4(true, x736, x737)∧COND4(gr(x737, 0), x736, p(x737))=COND4(true, x738, x739) ⇒ COND4(true, x738, x739)≥COND4(gr(x739, 0), x738, p(x739)))



    We simplified constraint (208) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (209)    (0=x1349gr(x735, x1349)=truex735=x7370=x1350gr(x737, x1350)=truep(x737)=x739COND4(true, x734, x739)≥COND4(gr(x739, 0), x734, p(x739)))



    We simplified constraint (209) using rule (V) (with possible (I) afterwards) using induction on gr(x735, x1349)=true which results in the following new constraints:

    (210)    (true=true0=0s(x1352)=x7370=x1350gr(x737, x1350)=truep(x737)=x739COND4(true, x734, x739)≥COND4(gr(x739, 0), x734, p(x739)))


    (211)    (gr(x1354, x1353)=true0=s(x1353)∧s(x1354)=x7370=x1350gr(x737, x1350)=truep(x737)=x739∧(∀x1355,x1356,x1357,x1358:gr(x1354, x1353)=true0=x1353x1354=x13550=x1356gr(x1355, x1356)=truep(x1355)=x1357COND4(true, x1358, x1357)≥COND4(gr(x1357, 0), x1358, p(x1357))) ⇒ COND4(true, x734, x739)≥COND4(gr(x739, 0), x734, p(x739)))



    We simplified constraint (210) using rules (I), (II) which results in the following new constraint:

    (212)    (s(x1352)=x7370=x1350gr(x737, x1350)=truep(x737)=x739COND4(true, x734, x739)≥COND4(gr(x739, 0), x734, p(x739)))



    We solved constraint (211) using rules (I), (II).We simplified constraint (212) using rule (V) (with possible (I) afterwards) using induction on gr(x737, x1350)=true which results in the following new constraints:

    (213)    (true=trues(x1352)=s(x1360)∧0=0p(s(x1360))=x739COND4(true, x734, x739)≥COND4(gr(x739, 0), x734, p(x739)))


    (214)    (gr(x1362, x1361)=trues(x1352)=s(x1362)∧0=s(x1361)∧p(s(x1362))=x739∧(∀x1363,x1364,x1365:gr(x1362, x1361)=trues(x1363)=x13620=x1361p(x1362)=x1364COND4(true, x1365, x1364)≥COND4(gr(x1364, 0), x1365, p(x1364))) ⇒ COND4(true, x734, x739)≥COND4(gr(x739, 0), x734, p(x739)))



    We simplified constraint (213) using rules (I), (II), (III), (IV), (VII) which results in the following new constraint:

    (215)    (COND4(true, x734, x739)≥COND4(gr(x739, 0), x734, p(x739)))



    We solved constraint (214) using rules (I), (II).
  • We consider the chain COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)) which results in the following constraint:

    (216)    (COND4(gr(x741, 0), x740, p(x741))=COND4(true, x742, x743)∧COND4(gr(x743, 0), x742, p(x743))=COND4(true, x744, x745) ⇒ COND4(true, x744, x745)≥COND4(gr(x745, 0), x744, p(x745)))



    We simplified constraint (216) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (217)    (0=x1367gr(x741, x1367)=truep(x741)=x7430=x1368gr(x743, x1368)=truep(x743)=x745COND4(true, x740, x745)≥COND4(gr(x745, 0), x740, p(x745)))



    We simplified constraint (217) using rule (V) (with possible (I) afterwards) using induction on gr(x741, x1367)=true which results in the following new constraints:

    (218)    (true=true0=0p(s(x1370))=x7430=x1368gr(x743, x1368)=truep(x743)=x745COND4(true, x740, x745)≥COND4(gr(x745, 0), x740, p(x745)))


    (219)    (gr(x1372, x1371)=true0=s(x1371)∧p(s(x1372))=x7430=x1368gr(x743, x1368)=truep(x743)=x745∧(∀x1373,x1374,x1375,x1376:gr(x1372, x1371)=true0=x1371p(x1372)=x13730=x1374gr(x1373, x1374)=truep(x1373)=x1375COND4(true, x1376, x1375)≥COND4(gr(x1375, 0), x1376, p(x1375))) ⇒ COND4(true, x740, x745)≥COND4(gr(x745, 0), x740, p(x745)))



    We simplified constraint (218) using rules (I), (II), (VII) which results in the following new constraint:

    (220)    (s(x1370)=x1377p(x1377)=x7430=x1368gr(x743, x1368)=truep(x743)=x745COND4(true, x740, x745)≥COND4(gr(x745, 0), x740, p(x745)))



    We solved constraint (219) using rules (I), (II).We simplified constraint (220) using rule (V) (with possible (I) afterwards) using induction on gr(x743, x1368)=true which results in the following new constraints:

    (221)    (true=trues(x1370)=x1377p(x1377)=s(x1379)∧0=0p(s(x1379))=x745COND4(true, x740, x745)≥COND4(gr(x745, 0), x740, p(x745)))


    (222)    (gr(x1381, x1380)=trues(x1370)=x1377p(x1377)=s(x1381)∧0=s(x1380)∧p(s(x1381))=x745∧(∀x1382,x1383,x1384,x1385:gr(x1381, x1380)=trues(x1382)=x1383p(x1383)=x13810=x1380p(x1381)=x1384COND4(true, x1385, x1384)≥COND4(gr(x1384, 0), x1385, p(x1384))) ⇒ COND4(true, x740, x745)≥COND4(gr(x745, 0), x740, p(x745)))



    We simplified constraint (221) using rules (I), (II), (VII) which results in the following new constraint:

    (223)    (s(x1370)=x1377p(x1377)=s(x1379)∧s(x1379)=x1386p(x1386)=x745COND4(true, x740, x745)≥COND4(gr(x745, 0), x740, p(x745)))



    We solved constraint (222) using rules (I), (II).We simplified constraint (223) using rule (V) (with possible (I) afterwards) using induction on p(x1377)=s(x1379) which results in the following new constraint:

    (224)    (x1387=s(x1379)∧s(x1370)=s(x1387)∧s(x1379)=x1386p(x1386)=x745COND4(true, x740, x745)≥COND4(gr(x745, 0), x740, p(x745)))



    We simplified constraint (224) using rules (I), (II), (III), (IV) which results in the following new constraint:

    (225)    (COND4(true, x740, x745)≥COND4(gr(x745, 0), x740, p(x745)))







For Pair COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) the following chains were created:
  • We consider the chain COND1(true, x, y) → COND2(gr(x, y), x, y), COND2(false, x, y) → COND4(gr(y, 0), x, y), COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:

    (226)    (COND2(gr(x842, x843), x842, x843)=COND2(false, x844, x845)∧COND4(gr(x845, 0), x844, x845)=COND4(false, x846, x847) ⇒ COND4(false, x846, x847)≥COND1(and(gr(x846, 0), gr(x847, 0)), x846, x847))



    We simplified constraint (226) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (227)    (gr(x842, x843)=falsex843=x8450=x1388gr(x845, x1388)=falseCOND4(false, x842, x845)≥COND1(and(gr(x842, 0), gr(x845, 0)), x842, x845))



    We simplified constraint (227) using rule (V) (with possible (I) afterwards) using induction on gr(x842, x843)=false which results in the following new constraints:

    (228)    (false=falsex1389=x8450=x1388gr(x845, x1388)=falseCOND4(false, 0, x845)≥COND1(and(gr(0, 0), gr(x845, 0)), 0, x845))


    (229)    (gr(x1392, x1391)=falses(x1391)=x8450=x1388gr(x845, x1388)=false∧(∀x1393,x1394:gr(x1392, x1391)=falsex1391=x13930=x1394gr(x1393, x1394)=falseCOND4(false, x1392, x1393)≥COND1(and(gr(x1392, 0), gr(x1393, 0)), x1392, x1393)) ⇒ COND4(false, s(x1392), x845)≥COND1(and(gr(s(x1392), 0), gr(x845, 0)), s(x1392), x845))



    We simplified constraint (228) using rules (I), (II), (III) which results in the following new constraint:

    (230)    (0=x1388gr(x845, x1388)=falseCOND4(false, 0, x845)≥COND1(and(gr(0, 0), gr(x845, 0)), 0, x845))



    We simplified constraint (229) using rule (V) (with possible (I) afterwards) using induction on gr(x845, x1388)=false which results in the following new constraints:

    (231)    (false=falsegr(x1392, x1391)=falses(x1391)=00=x1399∧(∀x1393,x1394:gr(x1392, x1391)=falsex1391=x13930=x1394gr(x1393, x1394)=falseCOND4(false, x1392, x1393)≥COND1(and(gr(x1392, 0), gr(x1393, 0)), x1392, x1393)) ⇒ COND4(false, s(x1392), 0)≥COND1(and(gr(s(x1392), 0), gr(0, 0)), s(x1392), 0))


    (232)    (gr(x1402, x1401)=falsegr(x1392, x1391)=falses(x1391)=s(x1402)∧0=s(x1401)∧(∀x1393,x1394:gr(x1392, x1391)=falsex1391=x13930=x1394gr(x1393, x1394)=falseCOND4(false, x1392, x1393)≥COND1(and(gr(x1392, 0), gr(x1393, 0)), x1392, x1393))∧(∀x1403,x1404,x1405,x1406:gr(x1402, x1401)=falsegr(x1403, x1404)=falses(x1404)=x14020=x1401∧(∀x1405,x1406:gr(x1403, x1404)=falsex1404=x14050=x1406gr(x1405, x1406)=falseCOND4(false, x1403, x1405)≥COND1(and(gr(x1403, 0), gr(x1405, 0)), x1403, x1405)) ⇒ COND4(false, s(x1403), x1402)≥COND1(and(gr(s(x1403), 0), gr(x1402, 0)), s(x1403), x1402)) ⇒ COND4(false, s(x1392), s(x1402))≥COND1(and(gr(s(x1392), 0), gr(s(x1402), 0)), s(x1392), s(x1402)))



    We simplified constraint (230) using rule (V) (with possible (I) afterwards) using induction on gr(x845, x1388)=false which results in the following new constraints:

    (233)    (false=false0=x1395COND4(false, 0, 0)≥COND1(and(gr(0, 0), gr(0, 0)), 0, 0))


    (234)    (gr(x1398, x1397)=false0=s(x1397)∧(gr(x1398, x1397)=false0=x1397COND4(false, 0, x1398)≥COND1(and(gr(0, 0), gr(x1398, 0)), 0, x1398)) ⇒ COND4(false, 0, s(x1398))≥COND1(and(gr(0, 0), gr(s(x1398), 0)), 0, s(x1398)))



    We simplified constraint (233) using rules (I), (II), (IV) which results in the following new constraint:

    (235)    (COND4(false, 0, 0)≥COND1(and(gr(0, 0), gr(0, 0)), 0, 0))



    We solved constraint (234) using rules (I), (II).We solved constraint (231) using rules (I), (II).We solved constraint (232) using rules (I), (II).
  • We consider the chain COND2(false, x, y) → COND4(gr(y, 0), x, y), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:

    (236)    (COND4(gr(x863, 0), x862, x863)=COND4(true, x864, x865)∧COND4(gr(x865, 0), x864, p(x865))=COND4(false, x866, x867) ⇒ COND4(false, x866, x867)≥COND1(and(gr(x866, 0), gr(x867, 0)), x866, x867))



    We simplified constraint (236) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (237)    (0=x1407gr(x863, x1407)=truex863=x8650=x1408gr(x865, x1408)=falsep(x865)=x867COND4(false, x862, x867)≥COND1(and(gr(x862, 0), gr(x867, 0)), x862, x867))



    We simplified constraint (237) using rule (V) (with possible (I) afterwards) using induction on gr(x863, x1407)=true which results in the following new constraints:

    (238)    (true=true0=0s(x1410)=x8650=x1408gr(x865, x1408)=falsep(x865)=x867COND4(false, x862, x867)≥COND1(and(gr(x862, 0), gr(x867, 0)), x862, x867))


    (239)    (gr(x1412, x1411)=true0=s(x1411)∧s(x1412)=x8650=x1408gr(x865, x1408)=falsep(x865)=x867∧(∀x1413,x1414,x1415,x1416:gr(x1412, x1411)=true0=x1411x1412=x14130=x1414gr(x1413, x1414)=falsep(x1413)=x1415COND4(false, x1416, x1415)≥COND1(and(gr(x1416, 0), gr(x1415, 0)), x1416, x1415)) ⇒ COND4(false, x862, x867)≥COND1(and(gr(x862, 0), gr(x867, 0)), x862, x867))



    We simplified constraint (238) using rules (I), (II) which results in the following new constraint:

    (240)    (s(x1410)=x8650=x1408gr(x865, x1408)=falsep(x865)=x867COND4(false, x862, x867)≥COND1(and(gr(x862, 0), gr(x867, 0)), x862, x867))



    We solved constraint (239) using rules (I), (II).We simplified constraint (240) using rule (V) (with possible (I) afterwards) using induction on gr(x865, x1408)=false which results in the following new constraints:

    (241)    (false=falses(x1410)=00=x1417p(0)=x867COND4(false, x862, x867)≥COND1(and(gr(x862, 0), gr(x867, 0)), x862, x867))


    (242)    (gr(x1420, x1419)=falses(x1410)=s(x1420)∧0=s(x1419)∧p(s(x1420))=x867∧(∀x1421,x1422,x1423:gr(x1420, x1419)=falses(x1421)=x14200=x1419p(x1420)=x1422COND4(false, x1423, x1422)≥COND1(and(gr(x1423, 0), gr(x1422, 0)), x1423, x1422)) ⇒ COND4(false, x862, x867)≥COND1(and(gr(x862, 0), gr(x867, 0)), x862, x867))



    We solved constraint (241) using rules (I), (II).We solved constraint (242) using rules (I), (II).
  • We consider the chain COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(true, x, y) → COND4(gr(y, 0), x, p(y)), COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y) which results in the following constraint:

    (243)    (COND4(gr(x869, 0), x868, p(x869))=COND4(true, x870, x871)∧COND4(gr(x871, 0), x870, p(x871))=COND4(false, x872, x873) ⇒ COND4(false, x872, x873)≥COND1(and(gr(x872, 0), gr(x873, 0)), x872, x873))



    We simplified constraint (243) using rules (I), (II), (III), (VII) which results in the following new constraint:

    (244)    (0=x1424gr(x869, x1424)=truep(x869)=x8710=x1425gr(x871, x1425)=falsep(x871)=x873COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))



    We simplified constraint (244) using rule (V) (with possible (I) afterwards) using induction on gr(x869, x1424)=true which results in the following new constraints:

    (245)    (true=true0=0p(s(x1427))=x8710=x1425gr(x871, x1425)=falsep(x871)=x873COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))


    (246)    (gr(x1429, x1428)=true0=s(x1428)∧p(s(x1429))=x8710=x1425gr(x871, x1425)=falsep(x871)=x873∧(∀x1430,x1431,x1432,x1433:gr(x1429, x1428)=true0=x1428p(x1429)=x14300=x1431gr(x1430, x1431)=falsep(x1430)=x1432COND4(false, x1433, x1432)≥COND1(and(gr(x1433, 0), gr(x1432, 0)), x1433, x1432)) ⇒ COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))



    We simplified constraint (245) using rules (I), (II), (VII) which results in the following new constraint:

    (247)    (s(x1427)=x1434p(x1434)=x8710=x1425gr(x871, x1425)=falsep(x871)=x873COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))



    We solved constraint (246) using rules (I), (II).We simplified constraint (247) using rule (V) (with possible (I) afterwards) using induction on gr(x871, x1425)=false which results in the following new constraints:

    (248)    (false=falses(x1427)=x1434p(x1434)=00=x1435p(0)=x873COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))


    (249)    (gr(x1438, x1437)=falses(x1427)=x1434p(x1434)=s(x1438)∧0=s(x1437)∧p(s(x1438))=x873∧(∀x1439,x1440,x1441,x1442:gr(x1438, x1437)=falses(x1439)=x1440p(x1440)=x14380=x1437p(x1438)=x1441COND4(false, x1442, x1441)≥COND1(and(gr(x1442, 0), gr(x1441, 0)), x1442, x1441)) ⇒ COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))



    We simplified constraint (248) using rules (I), (II), (IV), (VII) which results in the following new constraint:

    (250)    (s(x1427)=x1434p(x1434)=00=x1443p(x1443)=x873COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))



    We solved constraint (249) using rules (I), (II).We simplified constraint (250) using rule (V) (with possible (I) afterwards) using induction on p(x1434)=0 which results in the following new constraints:

    (251)    (0=0s(x1427)=00=x1443p(x1443)=x873COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))


    (252)    (x1444=0s(x1427)=s(x1444)∧0=x1443p(x1443)=x873COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))



    We solved constraint (251) using rules (I), (II).We simplified constraint (252) using rules (I), (II), (III), (IV) which results in the following new constraint:

    (253)    (COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))







To summarize, we get the following constraints P for the following pairs.
  • COND2(true, x, y) → COND3(gr(x, 0), x, y)
    • (COND2(true, s(s(x947)), s(0))≥COND3(gr(s(s(x947)), 0), s(s(x947)), s(0)))
    • (COND2(true, s(x949), s(x948))≥COND3(gr(s(x949), 0), s(x949), s(x948)) ⇒ COND2(true, s(s(x949)), s(s(x948)))≥COND3(gr(s(s(x949)), 0), s(s(x949)), s(s(x948))))
    • (COND2(true, s(s(x1003)), s(0))≥COND3(gr(s(s(x1003)), 0), s(s(x1003)), s(0)))
    • (COND2(true, s(x1005), s(x1004))≥COND3(gr(s(x1005), 0), s(x1005), s(x1004)) ⇒ COND2(true, s(s(x1005)), s(s(x1004)))≥COND3(gr(s(s(x1005)), 0), s(s(x1005)), s(s(x1004))))

  • COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
    • (COND3(true, s(x1008), 0)≥COND3(gr(s(x1008), 0), p(s(x1008)), 0))
    • (COND3(true, s(s(x1027)), s(0))≥COND3(gr(s(s(x1027)), 0), p(s(s(x1027))), s(0)))
    • (COND3(true, s(x1029), s(x1028))≥COND3(gr(s(x1029), 0), p(s(x1029)), s(x1028)) ⇒ COND3(true, s(s(x1029)), s(s(x1028)))≥COND3(gr(s(s(x1029)), 0), p(s(s(x1029))), s(s(x1028))))
    • (COND3(true, x148, x145)≥COND3(gr(x148, 0), p(x148), x145))
    • (COND3(true, x154, x151)≥COND3(gr(x154, 0), p(x154), x151))

  • COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
    • (COND3(false, x282, x279)≥COND1(and(gr(x282, 0), gr(x279, 0)), x282, x279))

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

  • COND2(false, x, y) → COND4(gr(y, 0), x, y)
    • (COND2(false, s(0), s(x1273))≥COND4(gr(s(x1273), 0), s(0), s(x1273)))
    • (COND2(false, s(x1276), s(x1275))≥COND4(gr(s(x1275), 0), s(x1276), s(x1275)) ⇒ COND2(false, s(s(x1276)), s(s(x1275)))≥COND4(gr(s(s(x1275)), 0), s(s(x1276)), s(s(x1275))))
    • (COND2(false, s(0), s(x1322))≥COND4(gr(s(x1322), 0), s(0), s(x1322)))
    • (COND2(false, s(x1325), s(x1324))≥COND4(gr(s(x1324), 0), s(x1325), s(x1324)) ⇒ COND2(false, s(s(x1325)), s(s(x1324)))≥COND4(gr(s(s(x1324)), 0), s(s(x1325)), s(s(x1324))))

  • COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
    • (COND4(true, 0, s(x1334))≥COND4(gr(s(x1334), 0), 0, p(s(x1334))))
    • (COND4(true, s(0), s(x1345))≥COND4(gr(s(x1345), 0), s(0), p(s(x1345))))
    • (COND4(true, s(x1348), s(x1347))≥COND4(gr(s(x1347), 0), s(x1348), p(s(x1347))) ⇒ COND4(true, s(s(x1348)), s(s(x1347)))≥COND4(gr(s(s(x1347)), 0), s(s(x1348)), p(s(s(x1347)))))
    • (COND4(true, x734, x739)≥COND4(gr(x739, 0), x734, p(x739)))
    • (COND4(true, x740, x745)≥COND4(gr(x745, 0), x740, p(x745)))

  • COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
    • (COND4(false, 0, 0)≥COND1(and(gr(0, 0), gr(0, 0)), 0, 0))
    • (COND4(false, x868, x873)≥COND1(and(gr(x868, 0), gr(x873, 0)), x868, x873))




The constraints for P> respective Pbound 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 Pbound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(COND1(x1, x2, x3)) = -1 + 2·x2·x3 + x22 + x32   
POL(COND2(x1, x2, x3)) = 1 - 2·x2·x3 + x22 + x32   
POL(COND3(x1, x2, x3)) = 2·x2·x3 + x22 + x32   
POL(COND4(x1, x2, x3)) = 2·x2·x3 + x22 + x32   
POL(and(x1, x2)) = 0   
POL(c) = -1   
POL(false) = 0   
POL(gr(x1, x2)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   
POL(true) = 0   

The following pairs are in P>:

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
The following pairs are in Pbound:

COND2(true, x, y) → COND3(gr(x, 0), x, y)
COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND3(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
COND1(true, x, y) → COND2(gr(x, y), x, y)
COND2(false, x, y) → COND4(gr(y, 0), x, y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
COND4(false, x, y) → COND1(and(gr(x, 0), gr(y, 0)), x, y)
The following rules are usable:

p(s(x)) → x
p(0) → 0

(20) Obligation:

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

COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
COND4(true, x, y) → COND4(gr(y, 0), x, p(y))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x
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))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(22) Complex Obligation (AND)

(23) Obligation:

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

COND4(true, x, y) → COND4(gr(y, 0), x, p(y))

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x
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))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

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

(24) 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.

(25) Obligation:

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

COND4(true, x, y) → COND4(gr(y, 0), x, p(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

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

(26) 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].

and(true, true)
and(false, x0)
and(x0, false)

(27) Obligation:

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

COND4(true, x, y) → COND4(gr(y, 0), x, p(y))

The TRS R consists of the following rules:

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

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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


COND4(true, x, y) → COND4(gr(y, 0), x, p(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(COND4(x1, x2, x3)) = (2)x1 + x3   
POL(true) = 1   
POL(gr(x1, x2)) = (1/4)x1   
POL(0) = 0   
POL(p(x1)) = (1/2)x1   
POL(false) = 0   
POL(s(x1)) = 4 + (2)x1   
The value of delta used in the strict ordering is 2.
The following usable rules [FROCOS05] were oriented:

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

(29) Obligation:

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

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

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) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE

(32) Obligation:

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

COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

gr(0, x) → false
gr(s(x), 0) → true
and(true, true) → true
and(false, x) → false
and(x, false) → false
p(0) → 0
p(s(x)) → x
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))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

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

(33) 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.

(34) Obligation:

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

COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

gr(0, x0)
gr(s(x0), 0)
gr(s(x0), s(x1))
and(true, true)
and(false, x0)
and(x0, false)
p(0)
p(s(x0))

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

(35) 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].

and(true, true)
and(false, x0)
and(x0, false)

(36) Obligation:

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

COND3(true, x, y) → COND3(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

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

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.

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


COND3(true, x, y) → COND3(gr(x, 0), p(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO,RATPOLO]:

POL(COND3(x1, x2, x3)) = x1 + (4)x2   
POL(true) = 4   
POL(gr(x1, x2)) = (2)x1   
POL(0) = 0   
POL(p(x1)) = (1/4)x1   
POL(s(x1)) = 2 + (4)x1   
POL(false) = 0   
The value of delta used in the strict ordering is 4.
The following usable rules [FROCOS05] were oriented:

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

(38) Obligation:

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

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

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.

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE