Termination w.r.t. Q proof of /tmp/tmpjsMbp6/Ex49_GM04

Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 PisEmptyProof (⇔)

    ↳9 TRUE

  ↳10 QDP

    ↳11 QDPOrderProof (⇔)

    ↳12 QDP

    ↳13 PisEmptyProof (⇔)

    ↳14 TRUE

  ↳15 QDP

    ↳16 QDPOrderProof (⇔)

    ↳17 QDP

    ↳18 DependencyGraphProof (⇔)

    ↳19 TRUE

(0) Obligation:

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)
A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__DIV(s(X), s(Y)) → A__GEQ(X, Y)
A__IF(true, X, Y) → MARK(X)
A__IF(false, X, Y) → MARK(Y)
MARK(minus(X1, X2)) → A__MINUS(X1, X2)
MARK(geq(X1, X2)) → A__GEQ(X1, X2)
MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A__GEQ(s(X), s(Y)) → A__GEQ(X, Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__GEQ(x1, x2) = x1
s(x1) = s(x1)
a__minus(x1, x2) = x1
0 = 0
a__geq(x1, x2) = a__geq(x1, x2)
true = true
false = false
a__div(x1, x2) = a__div(x1, x2)
a__if(x1, x2, x3) = a__if(x1, x2, x3)
div(x1, x2) = div(x1, x2)
minus(x1, x2) = x1
mark(x1) = x1
geq(x1, x2) = geq(x1, x2)
if(x1, x2, x3) = if(x1, x2, x3)

Lexicographic path order with status [LPO].
Quasi-Precedence:

[a_{_geq₂}, false, a_{_div₂}, a_{_if₃}, div₂, geq₂, if₃] > [s₁, 0, true]

Status:

a_{_geq₂}: [2,1]
if₃: [3,2,1]
div₂: [2,1]
true: []
false: []
geq₂: [2,1]
s₁: [1]
0: []
a_{_div₂}: [2,1]
a_{_if₃}: [3,2,1]

The following usable rules [FROCOS05] were oriented:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

(7) Obligation:

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A__MINUS(s(X), s(Y)) → A__MINUS(X, Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__MINUS(x1, x2) = x1
s(x1) = s(x1)
a__minus(x1, x2) = x1
0 = 0
a__geq(x1, x2) = a__geq(x1, x2)
true = true
false = false
a__div(x1, x2) = a__div(x1, x2)
a__if(x1, x2, x3) = a__if(x1, x2, x3)
div(x1, x2) = div(x1, x2)
minus(x1, x2) = x1
mark(x1) = x1
geq(x1, x2) = geq(x1, x2)
if(x1, x2, x3) = if(x1, x2, x3)

Lexicographic path order with status [LPO].
Quasi-Precedence:

[a_{_geq₂}, false, a_{_div₂}, a_{_if₃}, div₂, geq₂, if₃] > [s₁, 0, true]

Status:

a_{_geq₂}: [2,1]
if₃: [3,2,1]
div₂: [2,1]
true: []
false: []
geq₂: [2,1]
s₁: [1]
0: []
a_{_div₂}: [2,1]
a_{_if₃}: [3,2,1]

The following usable rules [FROCOS05] were oriented:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

(12) Obligation:

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

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

MARK(div(X1, X2)) → A__DIV(mark(X1), X2)
A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__IF(true, X, Y) → MARK(X)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(false, X, Y) → MARK(Y)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(16) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

A__DIV(s(X), s(Y)) → A__IF(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
A__IF(true, X, Y) → MARK(X)
MARK(div(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__IF(false, X, Y) → MARK(Y)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = x1
div(x1, x2) = div(x1, x2)
A__DIV(x1, x2) = A__DIV(x1, x2)
mark(x1) = x1
s(x1) = s(x1)
A__IF(x1, x2, x3) = A__IF(x2, x3)
a__geq(x1, x2) = a__geq
minus(x1, x2) = minus(x1)
0 = 0
true = true
if(x1, x2, x3) = if(x1, x2, x3)
false = false
a__minus(x1, x2) = a__minus(x1)
a__div(x1, x2) = a__div(x1, x2)
a__if(x1, x2, x3) = a__if(x1, x2, x3)
geq(x1, x2) = geq

Lexicographic path order with status [LPO].
Quasi-Precedence:

[div₂, A_{_DIV₂}, a_{_geq}, a_{_div₂}, geq] > 0 > [s₁, false] > [minus₁, a_{_minus₁}]
[div₂, A_{_DIV₂}, a_{_geq}, a_{_div₂}, geq] > 0 > [s₁, false] > [if₃, a_{_if₃}] > A_{_IF₂}
[div₂, A_{_DIV₂}, a_{_geq}, a_{_div₂}, geq] > true

Status:

a_{_geq}: []
true: []
geq: []
0: []
A_{_IF₂}: [1,2]
if₃: [1,3,2]
A_{_DIV₂}: [2,1]
div₂: [2,1]
minus₁: [1]
false: []
s₁: [1]
a_{_minus₁}: [1]
a_{_if₃}: [1,3,2]
a_{_div₂}: [2,1]

The following usable rules [FROCOS05] were oriented:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

(17) Obligation:

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

MARK(div(X1, X2)) → A__DIV(mark(X1), X2)

The TRS R consists of the following rules:

a__minus(0, Y) → 0
a__minus(s(X), s(Y)) → a__minus(X, Y)
a__geq(X, 0) → true
a__geq(0, s(Y)) → false
a__geq(s(X), s(Y)) → a__geq(X, Y)
a__div(0, s(Y)) → 0
a__div(s(X), s(Y)) → a__if(a__geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
mark(minus(X1, X2)) → a__minus(X1, X2)
mark(geq(X1, X2)) → a__geq(X1, X2)
mark(div(X1, X2)) → a__div(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(0) → 0
mark(s(X)) → s(mark(X))
mark(true) → true
mark(false) → false
a__minus(X1, X2) → minus(X1, X2)
a__geq(X1, X2) → geq(X1, X2)
a__div(X1, X2) → div(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)

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

(18) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) PisEmptyProof (EQUIVALENT transformation)

(9) TRUE

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) PisEmptyProof (EQUIVALENT transformation)

(14) TRUE

(15) Obligation:

(16) QDPOrderProof (EQUIVALENT transformation)

(17) Obligation:

(18) DependencyGraphProof (EQUIVALENT transformation)

(19) TRUE