Termination w.r.t. Q proof of /tmp/tmpZJKmcU/thiemann11.trs

Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS

↳1 Overlay + Local Confluence (⇔)

↳2 QTRS

↳3 DependencyPairsProof (⇔)

↳4 QDP

↳5 DependencyGraphProof (⇔)

↳6 AND

  ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 PisEmptyProof (⇔)

    ↳16 TRUE

  ↳17 QDP

    ↳18 QDPOrderProof (⇔)

    ↳19 QDP

    ↳20 PisEmptyProof (⇔)

    ↳21 TRUE

  ↳22 QDP

(0) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

LE(s(x), s(y)) → LE(x, y)
ID(s(x)) → ID(x)
MINUS(s(x), s(y)) → MINUS(x, y)
MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
MOD(x, y) → ZERO(x)
MOD(x, y) → ZERO(y)
MOD(x, y) → LE(y, x)
MOD(x, y) → ID(x)
MOD(x, y) → ID(y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))
IF3(true, x, y) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 6 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = MINUS(x1)
s(x1) = s(x1)
le(x1, x2) = le
0 = 0
true = true
false = false
zero(x1) = zero(x1)
id(x1) = x1
minus(x1, x2) = x1
mod(x1, x2) = x1
if_mod(x1, x2, x3, x4, x5) = x4
if2(x1, x2, x3, x4) = x3
if3(x1, x2, x3) = x2

Lexicographic Path Order [LPO].
Precedence:

MINUS₁ > 0
s₁ > 0
le > true > 0
le > false > 0
zero₁ > true > 0
zero₁ > false > 0

The following usable rules [FROCOS05] were oriented:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

(9) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

ID(s(x)) → ID(x)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ID(s(x)) → ID(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ID(x1) = ID(x1)
s(x1) = s(x1)
le(x1, x2) = le
0 = 0
true = true
false = false
zero(x1) = zero
id(x1) = x1
minus(x1, x2) = x1
mod(x1, x2) = x1
if_mod(x1, x2, x3, x4, x5) = x4
if2(x1, x2, x3, x4) = x3
if3(x1, x2, x3) = x2

Lexicographic Path Order [LPO].
Precedence:

s₁ > ID₁ > [0, false]
s₁ > [le, true, zero] > [0, false]

The following usable rules [FROCOS05] were oriented:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

(14) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

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

(18) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LE(x1, x2) = x1
s(x1) = s(x1)
le(x1, x2) = le
0 = 0
true = true
false = false
zero(x1) = zero
id(x1) = x1
minus(x1, x2) = x1
mod(x1, x2) = x1
if_mod(x1, x2, x3, x4, x5) = x4
if2(x1, x2, x3, x4) = x3
if3(x1, x2, x3) = x2

Lexicographic Path Order [LPO].
Precedence:

[le, true, zero] > s₁ > 0
[le, true, zero] > false > 0

The following usable rules [FROCOS05] were oriented:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

(19) Obligation:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

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

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) Obligation:

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

MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF3(true, x, y) → MOD(minus(x, y), s(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
zero(0)
zero(s(x0))
id(0)
id(s(x0))
minus(x0, 0)
minus(s(x0), s(x1))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, x0, x1, x2, x3)
if2(true, x0, x1, x2)
if2(false, x0, x1, x2)
if3(true, x0, x1)
if3(false, x0, x1)

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