Termination w.r.t. Q proof of /tmp/tmpM9FqGs/PEANO_nosorts

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

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 QDPOrderProof (⇔)

    ↳18 QDP

    ↳19 PisEmptyProof (⇔)

    ↳20 TRUE

  ↳21 QDP

    ↳22 QDPOrderProof (⇔)

    ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 PisEmptyProof (⇔)

    ↳27 TRUE

  ↳28 QDP

    ↳29 QDPOrderProof (⇔)

    ↳30 QDP

    ↳31 QDPOrderProof (⇔)

    ↳32 QDP

    ↳33 PisEmptyProof (⇔)

    ↳34 TRUE

  ↳35 QDP

    ↳36 QDPOrderProof (⇔)

    ↳37 QDP

    ↳38 QDPOrderProof (⇔)

    ↳39 QDP

    ↳40 PisEmptyProof (⇔)

    ↳41 TRUE

  ↳42 QDP

    ↳43 QDPOrderProof (⇔)

    ↳44 QDP

    ↳45 QDPOrderProof (⇔)

    ↳46 QDP

    ↳47 PisEmptyProof (⇔)

    ↳48 TRUE

(0) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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(plus(N, s(M))) → S(plus(N, M))
ACTIVE(plus(N, s(M))) → PLUS(N, M)
ACTIVE(and(X1, X2)) → AND(active(X1), X2)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
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)
ACTIVE(s(X)) → S(active(X))
ACTIVE(s(X)) → ACTIVE(X)
AND(mark(X1), X2) → AND(X1, X2)
PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
S(mark(X)) → S(X)
PROPER(and(X1, X2)) → AND(proper(X1), proper(X2))
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(plus(X1, X2)) → PLUS(proper(X1), proper(X2))
PROPER(plus(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X2)
PROPER(s(X)) → S(proper(X))
PROPER(s(X)) → PROPER(X)
AND(ok(X1), ok(X2)) → AND(X1, X2)
PLUS(ok(X1), ok(X2)) → PLUS(X1, X2)
S(ok(X)) → S(X)
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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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 6 SCCs with 11 less nodes.

(4) Complex Obligation (AND)

(5) 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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

S(mark(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) = x1
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x1, x2)
tt = tt
plus(x1, x2) = plus(x1, x2)
0 = 0
s(x1) = s(x1)
proper(x1) = x1
top(x1) = top(x1)

Lexicographic path order with status [LPO].
Precedence:

and₂ > mark₁
plus₂ > s₁ > mark₁

Status:

mark₁: [1]
and₂: [2,1]
tt: []
plus₂: [1,2]
0: []
s₁: [1]
top₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(7) Obligation:

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

S(ok(X)) → S(X)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

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)
active(x1) = active(x1)
and(x1, x2) = x2
tt = tt
mark(x1) = x1
plus(x1, x2) = x1
0 = 0
s(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

active₁ > ok₁
proper₁ > tt > ok₁
proper₁ > 0 > ok₁
top > ok₁

Status:

ok₁: [1]
active₁: [1]
tt: []
0: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(9) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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:

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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

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) = x2
mark(x1) = x1
ok(x1) = ok(x1)
active(x1) = x1
and(x1, x2) = x2
tt = tt
plus(x1, x2) = x1
0 = 0
s(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

proper₁ > ok₁
proper₁ > tt
proper₁ > 0

Status:

ok₁: [1]
tt: []
0: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(14) 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)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

PLUS(X1, mark(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) = x2
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x1, x2)
tt = tt
plus(x1, x2) = plus(x1, x2)
0 = 0
s(x1) = s(x1)
proper(x1) = x1
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

and₂ > mark₁
tt > ok > plus₂ > s₁ > mark₁
0 > ok > plus₂ > s₁ > mark₁
top > mark₁

Status:

mark₁: [1]
and₂: [1,2]
tt: []
plus₂: [1,2]
0: []
s₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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

(16) 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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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) 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, x2)
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x1, x2)
tt = tt
plus(x1, x2) = plus(x1, x2)
0 = 0
s(x1) = s(x1)
proper(x1) = x1
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

PLUS₂ > mark₁
and₂ > mark₁
tt > ok > plus₂ > s₁ > mark₁
0 > ok > plus₂ > s₁ > mark₁
top > mark₁

Status:

PLUS₂: [1,2]
mark₁: [1]
and₂: [1,2]
tt: []
plus₂: [1,2]
0: []
s₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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

(18) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) Obligation:

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

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

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

AND(ok(X1), ok(X2)) → AND(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2) = AND(x2)
ok(x1) = ok(x1)
mark(x1) = x1
active(x1) = active(x1)
and(x1, x2) = x2
tt = tt
plus(x1, x2) = x1
0 = 0
s(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

proper₁ > ok₁ > active₁ > AND₁
proper₁ > tt > AND₁
proper₁ > 0 > AND₁
top > active₁ > AND₁

Status:

AND₁: [1]
ok₁: [1]
active₁: [1]
tt: []
0: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(23) Obligation:

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

AND(mark(X1), X2) → AND(X1, X2)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(mark(X1), X2) → AND(X1, X2)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
AND(x1, x2) = AND(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x1, x2)
tt = tt
plus(x1, x2) = plus(x1, x2)
0 = 0
s(x1) = s(x1)
proper(x1) = x1
ok(x1) = ok
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

AND₂ > mark₁
and₂ > mark₁
tt > ok > plus₂ > s₁ > mark₁
0 > ok > plus₂ > s₁ > mark₁
top > mark₁

Status:

AND₂: [1,2]
mark₁: [1]
and₂: [1,2]
tt: []
plus₂: [1,2]
0: []
s₁: [1]
ok: []
top: []

The following usable rules [FROCOS05] were oriented:

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

(25) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE

(28) Obligation:

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

PROPER(and(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X2)
PROPER(s(X)) → PROPER(X)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

PROPER(and(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → PROPER(X1)
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
and(x1, x2) = and(x1, x2)
plus(x1, x2) = plus(x1, x2)
s(x1) = x1
active(x1) = x1
tt = tt
mark(x1) = x1
0 = 0
proper(x1) = x1
ok(x1) = x1
top(x1) = top(x1)

Lexicographic path order with status [LPO].
Precedence:

plus₂ > and₂
tt > and₂
0 > and₂
top₁ > and₂

Status:

and₂: [1,2]
plus₂: [1,2]
tt: []
0: []
top₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(30) 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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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) 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: Combined order from the following AFS and order.
PROPER(x1) = x1
s(x1) = s(x1)
active(x1) = x1
and(x1, x2) = x2
tt = tt
mark(x1) = x1
plus(x1, x2) = plus(x1, x2)
0 = 0
proper(x1) = x1
ok(x1) = x1
top(x1) = top(x1)

Lexicographic path order with status [LPO].
Precedence:

tt > s₁
plus₂ > s₁
0 > s₁
top₁ > s₁

Status:

s₁: [1]
tt: []
plus₂: [1,2]
0: []
top₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(32) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

(33) PisEmptyProof (EQUIVALENT transformation)

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

(34) TRUE

(35) Obligation:

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

ACTIVE(plus(X1, X2)) → ACTIVE(X1)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
ACTIVE(plus(X1, X2)) → ACTIVE(X2)
ACTIVE(s(X)) → ACTIVE(X)

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

ACTIVE(plus(X1, X2)) → ACTIVE(X1)
ACTIVE(and(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
plus(x1, x2) = plus(x1, x2)
and(x1, x2) = and(x1, x2)
s(x1) = x1
active(x1) = x1
tt = tt
mark(x1) = x1
0 = 0
proper(x1) = x1
ok(x1) = x1
top(x1) = top(x1)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

plus₂: [2,1]
and₂: [1,2]
tt: []
0: []
top₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(37) 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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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) 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: Combined order from the following AFS and order.
ACTIVE(x1) = x1
s(x1) = s(x1)
active(x1) = x1
and(x1, x2) = x2
tt = tt
mark(x1) = x1
plus(x1, x2) = plus(x1, x2)
0 = 0
proper(x1) = x1
ok(x1) = x1
top(x1) = top(x1)

Lexicographic path order with status [LPO].
Precedence:

tt > s₁
plus₂ > s₁
0 > s₁
top₁ > s₁

Status:

s₁: [1]
tt: []
plus₂: [1,2]
0: []
top₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(39) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

(40) PisEmptyProof (EQUIVALENT transformation)

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

(41) TRUE

(42) 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(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

TOP(mark(X)) → TOP(proper(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1) = x1
ok(x1) = x1
active(x1) = x1
mark(x1) = mark(x1)
proper(x1) = x1
and(x1, x2) = and(x1, x2)
tt = tt
plus(x1, x2) = plus(x1, x2)
0 = 0
s(x1) = s(x1)
top(x1) = top(x1)

Lexicographic path order with status [LPO].
Precedence:

and₂ > mark₁
plus₂ > s₁ > mark₁

Status:

mark₁: [1]
and₂: [1,2]
tt: []
plus₂: [2,1]
0: []
s₁: [1]
top₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(44) Obligation:

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

TOP(ok(X)) → TOP(active(X))

The TRS R consists of the following rules:

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

TOP(ok(X)) → TOP(active(X))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TOP(x1) = TOP(x1)
ok(x1) = ok(x1)
active(x1) = x1
and(x1, x2) = x2
tt = tt
mark(x1) = x1
plus(x1, x2) = x1
0 = 0
s(x1) = x1
proper(x1) = proper(x1)
top(x1) = top

Lexicographic path order with status [LPO].
Precedence:

TOP₁ > ok₁
proper₁ > tt > ok₁
proper₁ > 0 > ok₁
top > ok₁

Status:

TOP₁: [1]
ok₁: [1]
tt: []
0: []
proper₁: [1]
top: []

The following usable rules [FROCOS05] were oriented:

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

(46) Obligation:

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

active(and(tt, X)) → mark(X)
active(plus(N, 0)) → mark(N)
active(plus(N, s(M))) → mark(s(plus(N, M)))
active(and(X1, X2)) → and(active(X1), X2)
active(plus(X1, X2)) → plus(active(X1), X2)
active(plus(X1, X2)) → plus(X1, active(X2))
active(s(X)) → s(active(X))
and(mark(X1), X2) → mark(and(X1, X2))
plus(mark(X1), X2) → mark(plus(X1, X2))
plus(X1, mark(X2)) → mark(plus(X1, X2))
s(mark(X)) → mark(s(X))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(0) → ok(0)
proper(s(X)) → s(proper(X))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
s(ok(X)) → ok(s(X))
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.

(47) 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) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) QDPOrderProof (EQUIVALENT transformation)

(9) Obligation:

(10) PisEmptyProof (EQUIVALENT transformation)

(11) TRUE

(12) Obligation:

(13) QDPOrderProof (EQUIVALENT transformation)

(14) Obligation:

(15) QDPOrderProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) PisEmptyProof (EQUIVALENT transformation)

(20) TRUE

(21) Obligation:

(22) QDPOrderProof (EQUIVALENT transformation)

(23) Obligation:

(24) QDPOrderProof (EQUIVALENT transformation)

(25) Obligation:

(26) PisEmptyProof (EQUIVALENT transformation)

(27) TRUE

(28) Obligation:

(29) QDPOrderProof (EQUIVALENT transformation)

(30) Obligation:

(31) QDPOrderProof (EQUIVALENT transformation)

(32) Obligation:

(33) PisEmptyProof (EQUIVALENT transformation)

(34) TRUE

(35) Obligation:

(36) QDPOrderProof (EQUIVALENT transformation)

(37) Obligation:

(38) QDPOrderProof (EQUIVALENT transformation)

(39) Obligation:

(40) PisEmptyProof (EQUIVALENT transformation)

(41) TRUE

(42) Obligation:

(43) QDPOrderProof (EQUIVALENT transformation)

(44) Obligation:

(45) QDPOrderProof (EQUIVALENT transformation)

(46) Obligation:

(47) PisEmptyProof (EQUIVALENT transformation)

(48) TRUE