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

(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) 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.

(7) 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.

(8) 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.

(9) 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.

(10) 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.

(11) 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.