Termination w.r.t. Q proof of /tmp/tmp6OZmtS/PEANO_nosorts_noand

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

0 QTRS

↳1 DependencyPairsProof (⇔)

↳2 QDP

↳3 DependencyGraphProof (⇔)

↳4 AND

  ↳5 QDP

    ↳6 QDPOrderProof (⇔)

    ↳7 QDP

    ↳8 QDPOrderProof (⇔)

    ↳9 QDP

    ↳10 PisEmptyProof (⇔)

    ↳11 TRUE

  ↳12 QDP

    ↳13 QDPOrderProof (⇔)

    ↳14 QDP

    ↳15 QDPOrderProof (⇔)

    ↳16 QDP

    ↳17 PisEmptyProof (⇔)

    ↳18 TRUE

  ↳19 QDP

    ↳20 QDPOrderProof (⇔)

    ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 PisEmptyProof (⇔)

    ↳25 TRUE

  ↳26 QDP

    ↳27 QDPOrderProof (⇔)

    ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 PisEmptyProof (⇔)

    ↳32 TRUE

  ↳33 QDP

    ↳34 QDPOrderProof (⇔)

    ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 PisEmptyProof (⇔)

    ↳39 TRUE

  ↳40 QDP

    ↳41 QDPOrderProof (⇔)

    ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 PisEmptyProof (⇔)

    ↳46 TRUE

  ↳47 QDP

(0) Obligation:

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

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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:

ACTIVE(U11(tt, M, N)) → U12¹(tt, M, N)
ACTIVE(U12(tt, M, N)) → S(plus(N, M))
ACTIVE(U12(tt, M, N)) → PLUS(N, M)
ACTIVE(plus(N, s(M))) → U11¹(tt, M, N)
ACTIVE(U11(X1, X2, X3)) → U11¹(active(X1), X2, X3)
ACTIVE(U11(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(U12(X1, X2, X3)) → U12¹(active(X1), X2, X3)
ACTIVE(U12(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(plus(X1, X2)) → PLUS(active(X1), X2)
ACTIVE(plus(X1, X2)) → ACTIVE(X1)
ACTIVE(plus(X1, X2)) → PLUS(X1, active(X2))
ACTIVE(plus(X1, X2)) → ACTIVE(X2)
U11¹(mark(X1), X2, X3) → U11¹(X1, X2, X3)
U12¹(mark(X1), X2, X3) → U12¹(X1, X2, X3)
S(mark(X)) → S(X)
PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
PROPER(U11(X1, X2, X3)) → U11¹(proper(X1), proper(X2), proper(X3))
PROPER(U11(X1, X2, X3)) → PROPER(X1)
PROPER(U11(X1, X2, X3)) → PROPER(X2)
PROPER(U11(X1, X2, X3)) → PROPER(X3)
PROPER(U12(X1, X2, X3)) → U12¹(proper(X1), proper(X2), proper(X3))
PROPER(U12(X1, X2, X3)) → PROPER(X1)
PROPER(U12(X1, X2, X3)) → PROPER(X2)
PROPER(U12(X1, X2, X3)) → PROPER(X3)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
PROPER(plus(X1, X2)) → PLUS(proper(X1), proper(X2))
PROPER(plus(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X2)
U11¹(ok(X1), ok(X2), ok(X3)) → U11¹(X1, X2, X3)
U12¹(ok(X1), ok(X2), ok(X3)) → U12¹(X1, X2, X3)
S(ok(X)) → S(X)
PLUS(ok(X1), ok(X2)) → PLUS(X1, X2)
TOP(mark(X)) → TOP(proper(X))
TOP(mark(X)) → PROPER(X)
TOP(ok(X)) → TOP(active(X))
TOP(ok(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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 7 SCCs with 15 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

PLUS(X1, mark(X2)) → PLUS(X1, X2)
PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(ok(X1), ok(X2)) → PLUS(X1, X2)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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.

PLUS(X1, mark(X2)) → PLUS(X1, X2)
PLUS(ok(X1), ok(X2)) → PLUS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

ok₁ > PLUS₁

Status:

PLUS₁: [1]
mark₁: [1]
ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

PLUS(mark(X1), X2) → PLUS(X1, X2)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
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.

PLUS(mark(X1), X2) → PLUS(X1, X2)

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

Lexicographic path order with status [LPO].
Precedence:

mark₁ > PLUS₁

Status:

PLUS₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

Q is empty.
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:

S(ok(X)) → S(X)
S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(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.

S(ok(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
S(x1) = x1
ok(x1) = ok(x1)
mark(x1) = x1

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

ok₁: [1]

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

S(mark(X)) → S(X)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

S(mark(X)) → S(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > S₁

Status:

S₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) Obligation:

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

U12¹(ok(X1), ok(X2), ok(X3)) → U12¹(X1, X2, X3)
U12¹(mark(X1), X2, X3) → U12¹(X1, X2, X3)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(20) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U12¹(ok(X1), ok(X2), ok(X3)) → U12¹(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U12¹(x1, x2, x3) = U12¹(x3)
ok(x1) = ok(x1)
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

mark₁ > U12^1₁

Status:

U12^1₁: [1]
ok₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

U12¹(mark(X1), X2, X3) → U12¹(X1, X2, X3)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U12¹(mark(X1), X2, X3) → U12¹(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > U12^1₃

Status:

U12^1₃: [3,1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE

(26) Obligation:

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

U11¹(ok(X1), ok(X2), ok(X3)) → U11¹(X1, X2, X3)
U11¹(mark(X1), X2, X3) → U11¹(X1, X2, X3)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U11¹(ok(X1), ok(X2), ok(X3)) → U11¹(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U11¹(x1, x2, x3) = U11¹(x3)
ok(x1) = ok(x1)
mark(x1) = mark(x1)

Lexicographic path order with status [LPO].
Precedence:

mark₁ > U11^1₁

Status:

U11^1₁: [1]
ok₁: [1]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

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

U11¹(mark(X1), X2, X3) → U11¹(X1, X2, X3)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(29) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

U11¹(mark(X1), X2, X3) → U11¹(X1, X2, X3)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

mark₁ > U11^1₃

Status:

U11^1₃: [3,1,2]
mark₁: [1]

The following usable rules [FROCOS05] were oriented: none

(30) Obligation:

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

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(31) PisEmptyProof (EQUIVALENT transformation)

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

(32) TRUE

(33) Obligation:

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

PROPER(U11(X1, X2, X3)) → PROPER(X2)
PROPER(U11(X1, X2, X3)) → PROPER(X1)
PROPER(U11(X1, X2, X3)) → PROPER(X3)
PROPER(U12(X1, X2, X3)) → PROPER(X1)
PROPER(U12(X1, X2, X3)) → PROPER(X2)
PROPER(U12(X1, X2, X3)) → PROPER(X3)
PROPER(s(X)) → PROPER(X)
PROPER(plus(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X2)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(34) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(U11(X1, X2, X3)) → PROPER(X2)
PROPER(U11(X1, X2, X3)) → PROPER(X1)
PROPER(U11(X1, X2, X3)) → PROPER(X3)
PROPER(U12(X1, X2, X3)) → PROPER(X1)
PROPER(U12(X1, X2, X3)) → PROPER(X2)
PROPER(U12(X1, X2, X3)) → PROPER(X3)
PROPER(plus(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROPER(x1) = x1
U11(x1, x2, x3) = U11(x1, x2, x3)
U12(x1, x2, x3) = U12(x1, x2, x3)
s(x1) = x1
plus(x1, x2) = plus(x1, x2)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

U11₃: [1,2,3]
U12₃: [1,2,3]
plus₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

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

PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(36) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROPER(s(X)) → PROPER(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

s₁ > PROPER₁

Status:

PROPER₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(37) Obligation:

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

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(38) PisEmptyProof (EQUIVALENT transformation)

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

(39) TRUE

(40) Obligation:

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

ACTIVE(U12(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(U11(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(plus(X1, X2)) → ACTIVE(X1)
ACTIVE(plus(X1, X2)) → ACTIVE(X2)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(41) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(U12(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(U11(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(plus(X1, X2)) → ACTIVE(X1)
ACTIVE(plus(X1, X2)) → ACTIVE(X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACTIVE(x1) = x1
U12(x1, x2, x3) = U12(x1)
U11(x1, x2, x3) = U11(x1)
s(x1) = x1
plus(x1, x2) = plus(x1, x2)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

U12₁: [1]
U11₁: [1]
plus₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(42) Obligation:

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

ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(s(X)) → ACTIVE(X)

The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:

s₁ > ACTIVE₁

Status:

ACTIVE₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(44) Obligation:

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

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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

(45) PisEmptyProof (EQUIVALENT transformation)

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

(46) TRUE

(47) Obligation:

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

TOP(ok(X)) → TOP(active(X))
TOP(mark(X)) → TOP(proper(X))

The TRS R consists of the following rules:

active(U11(tt, M, N)) → mark(U12(tt, M, N))
active(U12(tt, M, N)) → mark(s(plus(N, M)))
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(U11(tt, M, N))
active(U11(X1, X2, X3)) → U11(active(X1), X2, X3)
active(U12(X1, X2, X3)) → U12(active(X1), X2, X3)
active(s(X)) → s(active(X))
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
U11(mark(X1), X2, X3) → mark(U11(X1, X2, X3))
U12(mark(X1), X2, X3) → mark(U12(X1, X2, X3))
s(mark(X)) → mark(s(X))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
proper(U11(X1, X2, X3)) → U11(proper(X1), proper(X2), proper(X3))
proper(tt) → ok(tt)
proper(U12(X1, X2, X3)) → U12(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
U11(ok(X1), ok(X2), ok(X3)) → ok(U11(X1, X2, X3))
U12(ok(X1), ok(X2), ok(X3)) → ok(U12(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))

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