Termination w.r.t. Q proof of /tmp/tmpDI4C4S/Ex2_Luc03b

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 QDP

↳5 QDPOrderProof (⇔)

↳6 QDP

↳7 QDPOrderProof (⇔)

↳8 QDP

↳9 QDPOrderProof (⇔)

↳10 QDP

↳11 DependencyGraphProof (⇔)

↳12 QDP

↳13 QDPOrderProof (⇔)

↳14 QDP

↳15 PisEmptyProof (⇔)

↳16 TRUE

(0) Obligation:

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

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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__FST(s(X), cons(Y, Z)) → MARK(Y)
A__FROM(X) → MARK(X)
A__ADD(0, X) → MARK(X)
MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → A__LEN(mark(X))
MARK(len(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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 1 SCC with 1 less node.

(4) Obligation:

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

MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
A__ADD(0, X) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)
MARK(len(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(fst(X1, X2)) → A__FST(mark(X1), mark(X2))
A__FST(s(X), cons(Y, Z)) → MARK(Y)
MARK(fst(X1, X2)) → MARK(X1)
MARK(fst(X1, X2)) → MARK(X2)
MARK(add(X1, X2)) → A__ADD(mark(X1), mark(X2))
A__ADD(0, X) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)
MARK(add(X1, X2)) → MARK(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
fst(x1, x2) = fst(x1, x2)
A__FST(x1, x2) = A__FST(x1, x2)
mark(x1) = x1
s(x1) = s
cons(x1, x2) = x1
from(x1) = x1
A__FROM(x1) = A__FROM(x1)
add(x1, x2) = add(x1, x2)
A__ADD(x1, x2) = A__ADD(x2)
0 = 0
len(x1) = x1
a__len(x1) = x1
a__add(x1, x2) = a__add(x1, x2)
a__from(x1) = x1
a__fst(x1, x2) = a__fst(x1, x2)
nil = nil

Recursive Path Order [RPO].
Precedence:

[fst₂, a_{_fst₂}] > [MARK₁, A_{_FST₂}, A_{_FROM₁}] > s
[fst₂, a_{_fst₂}] > nil > 0 > s
[add₂, A_{_ADD₁}, a_{_add₂}] > [MARK₁, A_{_FST₂}, A_{_FROM₁}] > s

The following usable rules [FROCOS05] were oriented:

a__len(X) → len(X)
a__add(X1, X2) → add(X1, X2)
a__from(X) → from(X)
a__fst(X1, X2) → fst(X1, X2)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(nil) → nil
mark(s(X)) → s(X)
mark(0) → 0
mark(len(X)) → a__len(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__add(0, X) → mark(X)
mark(from(X)) → a__from(mark(X))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
a__len(cons(X, Z)) → s(len(Z))
a__len(nil) → 0
a__add(s(X), Y) → s(add(X, Y))
a__from(X) → cons(mark(X), from(s(X)))
a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))

(6) Obligation:

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

MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(len(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(len(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
from(x1) = x1
A__FROM(x1) = A__FROM(x1)
mark(x1) = x1
len(x1) = len(x1)
cons(x1, x2) = x1
a__len(x1) = a__len(x1)
a__add(x1, x2) = x2
add(x1, x2) = x2
a__from(x1) = x1
a__fst(x1, x2) = x2
fst(x1, x2) = x2
nil = nil
s(x1) = s
0 = 0

Recursive Path Order [RPO].
Precedence:

[MARK₁, A_{_FROM₁}, len₁, a_{_len₁}, 0] > [nil, s]

The following usable rules [FROCOS05] were oriented:

a__len(X) → len(X)
a__add(X1, X2) → add(X1, X2)
a__from(X) → from(X)
a__fst(X1, X2) → fst(X1, X2)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(nil) → nil
mark(s(X)) → s(X)
mark(0) → 0
mark(len(X)) → a__len(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__add(0, X) → mark(X)
mark(from(X)) → a__from(mark(X))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
a__len(cons(X, Z)) → s(len(Z))
a__len(nil) → 0
a__add(s(X), Y) → s(add(X, Y))
a__from(X) → cons(mark(X), from(s(X)))
a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))

(8) Obligation:

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

MARK(from(X)) → A__FROM(mark(X))
A__FROM(X) → MARK(X)
MARK(from(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(from(X)) → A__FROM(mark(X))
MARK(from(X)) → MARK(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = MARK(x1)
from(x1) = from(x1)
A__FROM(x1) = A__FROM(x1)
mark(x1) = x1
cons(x1, x2) = x1
a__len(x1) = a__len
len(x1) = len
a__add(x1, x2) = x2
add(x1, x2) = x2
a__from(x1) = a__from(x1)
a__fst(x1, x2) = x2
fst(x1, x2) = x2
nil = nil
s(x1) = x1
0 = 0

Recursive Path Order [RPO].
Precedence:

[a_{_len}, len] > 0 > [MARK₁, from₁, A_{_FROM₁}, a_{_from₁}, nil]

The following usable rules [FROCOS05] were oriented:

a__len(X) → len(X)
a__add(X1, X2) → add(X1, X2)
a__from(X) → from(X)
a__fst(X1, X2) → fst(X1, X2)
mark(cons(X1, X2)) → cons(mark(X1), X2)
mark(nil) → nil
mark(s(X)) → s(X)
mark(0) → 0
mark(len(X)) → a__len(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
a__add(0, X) → mark(X)
mark(from(X)) → a__from(mark(X))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
a__len(cons(X, Z)) → s(len(Z))
a__len(nil) → 0
a__add(s(X), Y) → s(add(X, Y))
a__from(X) → cons(mark(X), from(s(X)))
a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))

(10) Obligation:

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

A__FROM(X) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

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

(11) DependencyGraphProof (EQUIVALENT transformation)

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

(12) Obligation:

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

MARK(cons(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

Q is empty.
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.

MARK(cons(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MARK(x1) = x1
cons(x1, x2) = cons(x1, x2)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

a__fst(0, Z) → nil
a__fst(s(X), cons(Y, Z)) → cons(mark(Y), fst(X, Z))
a__from(X) → cons(mark(X), from(s(X)))
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__len(nil) → 0
a__len(cons(X, Z)) → s(len(Z))
mark(fst(X1, X2)) → a__fst(mark(X1), mark(X2))
mark(from(X)) → a__from(mark(X))
mark(add(X1, X2)) → a__add(mark(X1), mark(X2))
mark(len(X)) → a__len(mark(X))
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(mark(X1), X2)
a__fst(X1, X2) → fst(X1, X2)
a__from(X) → from(X)
a__add(X1, X2) → add(X1, X2)
a__len(X) → len(X)

Q is empty.
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.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) QDPOrderProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) DependencyGraphProof (EQUIVALENT transformation)

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) PisEmptyProof (EQUIVALENT transformation)

(16) TRUE