Termination w.r.t. Q proof of /tmp/tmp0oj5O1/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 DependencyGraphProof (⇔)

↳8 QDP

↳9 QDPOrderProof (⇔)

↳10 QDP

↳11 QDPOrderProof (⇔)

↳12 QDP

↳13 DependencyGraphProof (⇔)

↳14 TRUE

(0) Obligation:

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

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → 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:

FST(s(X), cons(Y, Z)) → ACTIVATE(X)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)
ADD(s(X), Y) → S(n__add(activate(X), Y))
ADD(s(X), Y) → ACTIVATE(X)
LEN(cons(X, Z)) → S(n__len(activate(Z)))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → FROM(activate(X))
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__len(X)) → LEN(activate(X))
ACTIVATE(n__len(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → 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 4 less nodes.

(4) Obligation:

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

ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → ACTIVATE(X)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → 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.

ACTIVATE(n__fst(X1, X2)) → FST(activate(X1), activate(X2))
FST(s(X), cons(Y, Z)) → ACTIVATE(X)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fst(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__add(X1, X2)) → ADD(activate(X1), activate(X2))
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__add(X1, X2)) → ACTIVATE(X2)
FST(s(X), cons(Y, Z)) → ACTIVATE(Z)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__fst(x1, x2) = n__fst(x1, x2)
FST(x1, x2) = FST(x1, x2)
activate(x1) = x1
s(x1) = x1
cons(x1, x2) = x2
n__from(x1) = x1
n__add(x1, x2) = n__add(x1, x2)
ADD(x1, x2) = x1
n__len(x1) = x1
LEN(x1) = x1
fst(x1, x2) = fst(x1, x2)
0 = 0
nil = nil
add(x1, x2) = add(x1, x2)
from(x1) = x1
n__s(x1) = x1
len(x1) = x1

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

[n_{_fst₂}, fst₂] > FST₂
[n_{_fst₂}, fst₂] > nil > 0
[n_{_add₂}, add₂]

Status:

add₂: [1,2]
n_{_add₂}: [1,2]
n_{_fst₂}: [2,1]
fst₂: [2,1]

The following usable rules [FROCOS05] were oriented:

fst(0, Z) → nil
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
activate(n__len(X)) → len(activate(X))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__s(X)) → s(X)
activate(n__from(X)) → from(activate(X))
activate(X) → X

(6) Obligation:

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

ACTIVATE(n__from(X)) → ACTIVATE(X)
ADD(s(X), Y) → ACTIVATE(X)
ACTIVATE(n__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

ACTIVATE(n__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__from(X)) → ACTIVATE(X)
ACTIVATE(n__len(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → 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.

ACTIVATE(n__from(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__len(x1) = x1
LEN(x1) = x1
activate(x1) = x1
cons(x1, x2) = x2
n__from(x1) = n__from(x1)
fst(x1, x2) = fst
nil = nil
add(x1, x2) = x2
s(x1) = s
n__fst(x1, x2) = n__fst
from(x1) = from(x1)
n__s(x1) = n__s
len(x1) = x1
0 = 0
n__add(x1, x2) = x2

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

[n_{_from₁}, from₁] > [s, n_{_s}]
[fst, n_{_fst}] > nil > 0 > [s, n_{_s}]

Status:

trivial

The following usable rules [FROCOS05] were oriented:

fst(0, Z) → nil
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
activate(n__len(X)) → len(activate(X))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__s(X)) → s(X)
activate(n__from(X)) → from(activate(X))
activate(X) → X

(10) Obligation:

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

ACTIVATE(n__len(X)) → LEN(activate(X))
LEN(cons(X, Z)) → ACTIVATE(Z)
ACTIVATE(n__len(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X

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.

ACTIVATE(n__len(X)) → LEN(activate(X))
ACTIVATE(n__len(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1) = x1
n__len(x1) = n__len(x1)
LEN(x1) = x1
activate(x1) = x1
cons(x1, x2) = x2
fst(x1, x2) = fst
nil = nil
add(x1, x2) = x2
s(x1) = s
n__fst(x1, x2) = n__fst
from(x1) = from
n__from(x1) = n__from
len(x1) = len(x1)
0 = 0
n__s(x1) = n__s
n__add(x1, x2) = x2

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

[n_{_len₁}, len₁] > 0 > [s, n_{_s}]
[fst, n_{_fst}] > nil > [s, n_{_s}]
[from, n_{_from}] > [s, n_{_s}]

Status:

trivial

The following usable rules [FROCOS05] were oriented:

fst(0, Z) → nil
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
activate(n__len(X)) → len(activate(X))
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__s(X)) → s(X)
activate(n__from(X)) → from(activate(X))
activate(X) → X

(12) Obligation:

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

LEN(cons(X, Z)) → ACTIVATE(Z)

The TRS R consists of the following rules:

fst(0, Z) → nil
fst(s(X), cons(Y, Z)) → cons(Y, n__fst(activate(X), activate(Z)))
from(X) → cons(X, n__from(n__s(X)))
add(0, X) → X
add(s(X), Y) → s(n__add(activate(X), Y))
len(nil) → 0
len(cons(X, Z)) → s(n__len(activate(Z)))
fst(X1, X2) → n__fst(X1, X2)
from(X) → n__from(X)
s(X) → n__s(X)
add(X1, X2) → n__add(X1, X2)
len(X) → n__len(X)
activate(n__fst(X1, X2)) → fst(activate(X1), activate(X2))
activate(n__from(X)) → from(activate(X))
activate(n__s(X)) → s(X)
activate(n__add(X1, X2)) → add(activate(X1), activate(X2))
activate(n__len(X)) → len(activate(X))
activate(X) → X

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

(13) 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) Obligation:

(5) QDPOrderProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) DependencyGraphProof (EQUIVALENT transformation)

(14) TRUE