Termination w.r.t. Q proof of /tmp/tmpojJs98/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 QDPOrderProof (⇔)

    ↳20 QDP

    ↳21 PisEmptyProof (⇔)

    ↳22 TRUE

  ↳23 QDP

    ↳24 QDPOrderProof (⇔)

    ↳25 QDP

    ↳26 QDPOrderProof (⇔)

    ↳27 QDP

    ↳28 QDPOrderProof (⇔)

    ↳29 QDP

    ↳30 QDPOrderProof (⇔)

    ↳31 QDP

    ↳32 PisEmptyProof (⇔)

    ↳33 TRUE

  ↳34 QDP

    ↳35 QDPOrderProof (⇔)

    ↳36 QDP

    ↳37 QDPOrderProof (⇔)

    ↳38 QDP

    ↳39 PisEmptyProof (⇔)

    ↳40 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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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(and(tt, X)) → MARK(X)
ACTIVE(plus(N, 0)) → MARK(N)
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
ACTIVE(plus(N, s(M))) → S(plus(N, M))
ACTIVE(plus(N, s(M))) → PLUS(N, M)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(and(X1, X2)) → AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(tt) → ACTIVE(tt)
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(plus(X1, X2)) → PLUS(mark(X1), mark(X2))
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
MARK(0) → ACTIVE(0)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(s(X)) → S(mark(X))
MARK(s(X)) → MARK(X)
AND(mark(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
PLUS(active(X1), X2) → PLUS(X1, X2)
PLUS(X1, active(X2)) → PLUS(X1, X2)
S(mark(X)) → S(X)
S(active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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 4 SCCs with 7 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

S(active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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
active(x1) = x1
mark(x1) = mark(x1)
and(x1, x2) = and(x2)
tt = tt
plus(x1, x2) = plus(x1)
0 = 0
s(x1) = s

Lexicographic Path Order [LPO].
Precedence:

and₁ > [mark₁, plus₁] > 0 > s
tt > [mark₁, plus₁] > 0 > s

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(7) Obligation:

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

S(active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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(active(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
active(x1) = active(x1)
and(x1, x2) = and(x2)
tt = tt
mark(x1) = mark(x1)
plus(x1, x2) = plus(x1, x2)
0 = 0
s(x1) = s

Lexicographic Path Order [LPO].
Precedence:

[and₁, tt, mark₁, plus₂] > active₁ > s
[and₁, tt, mark₁, plus₂] > 0 > s

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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(active(X1), X2) → PLUS(X1, X2)
PLUS(X1, active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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(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) = x1
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x2)
tt = tt
plus(x1, x2) = plus(x1)
0 = 0
s(x1) = s

Lexicographic Path Order [LPO].
Precedence:

0 > [mark₁, and₁, plus₁] > tt > s

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(14) Obligation:

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

PLUS(X1, mark(X2)) → PLUS(X1, X2)
PLUS(active(X1), X2) → PLUS(X1, X2)
PLUS(X1, active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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) = PLUS(x1, x2)
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x2)
tt = tt
plus(x1, x2) = plus(x1)
0 = 0
s(x1) = s

Lexicographic Path Order [LPO].
Precedence:

and₁ > [PLUS₂, mark₁, plus₁] > tt > s
and₁ > [PLUS₂, mark₁, plus₁] > 0 > s

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(16) Obligation:

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

PLUS(active(X1), X2) → PLUS(X1, X2)
PLUS(X1, active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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(X1, active(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
active(x1) = active(x1)
and(x1, x2) = and(x2)
tt = tt
mark(x1) = mark(x1)
plus(x1, x2) = plus(x1, x2)
0 = 0
s(x1) = s

Lexicographic Path Order [LPO].
Precedence:

and₁ > [mark₁, plus₂] > [active₁, s]
tt > [mark₁, plus₂] > [active₁, s]
0 > [mark₁, plus₂] > [active₁, s]

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(18) Obligation:

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

PLUS(active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(19) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

Lexicographic Path Order [LPO].
Precedence:

[and₁, tt, mark₁, plus₂, 0] > active₁ > s

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(20) 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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(21) PisEmptyProof (EQUIVALENT transformation)

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

(22) TRUE

(23) Obligation:

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

AND(X1, mark(X2)) → AND(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(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) = x1
mark(x1) = mark(x1)
active(x1) = x1
and(x1, x2) = and(x2)
tt = tt
plus(x1, x2) = plus(x1)
0 = 0
s(x1) = s

Lexicographic Path Order [LPO].
Precedence:

0 > [mark₁, and₁, plus₁] > tt > s

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(25) Obligation:

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

AND(X1, mark(X2)) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(X1, mark(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(x2)
tt = tt
plus(x1, x2) = plus(x1)
0 = 0
s(x1) = s

Lexicographic Path Order [LPO].
Precedence:

and₁ > [AND₂, mark₁, plus₁] > tt > s
and₁ > [AND₂, mark₁, plus₁] > 0 > s

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(27) Obligation:

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

AND(active(X1), X2) → AND(X1, X2)
AND(X1, active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(28) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

AND(X1, active(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) = x2
active(x1) = active(x1)
and(x1, x2) = and(x2)
tt = tt
mark(x1) = mark(x1)
plus(x1, x2) = plus(x1, x2)
0 = 0
s(x1) = s

Lexicographic Path Order [LPO].
Precedence:

and₁ > [mark₁, plus₂] > [active₁, s]
tt > [mark₁, plus₂] > [active₁, s]
0 > [mark₁, plus₂] > [active₁, s]

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(29) Obligation:

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

AND(active(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

Lexicographic Path Order [LPO].
Precedence:

[and₁, tt, mark₁, plus₂, 0] > active₁ > s

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(31) 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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(32) PisEmptyProof (EQUIVALENT transformation)

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

(33) TRUE

(34) Obligation:

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

MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
ACTIVE(and(tt, X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
ACTIVE(plus(N, 0)) → MARK(N)
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
MARK(s(X)) → MARK(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(35) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACTIVE(and(tt, X)) → MARK(X)
MARK(and(X1, X2)) → MARK(X1)
ACTIVE(plus(N, 0)) → MARK(N)
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)

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

Lexicographic Path Order [LPO].
Precedence:

trivial

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(36) Obligation:

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

MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
MARK(s(X)) → MARK(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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(37) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(plus(X1, X2)) → ACTIVE(plus(mark(X1), mark(X2)))
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(plus(N, s(M))) → MARK(s(plus(N, M)))
MARK(s(X)) → MARK(X)

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

Lexicographic Path Order [LPO].
Precedence:

and₂ > ACTIVE₁
plus₂ > [MARK₁, s₁] > ACTIVE₁
tt > ACTIVE₁
0 > ACTIVE₁

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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

(38) 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)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(plus(X1, X2)) → active(plus(mark(X1), mark(X2)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
plus(mark(X1), X2) → plus(X1, X2)
plus(X1, mark(X2)) → plus(X1, X2)
plus(active(X1), X2) → plus(X1, X2)
plus(X1, active(X2)) → plus(X1, X2)
s(mark(X)) → s(X)
s(active(X)) → s(X)

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

(39) 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) QDPOrderProof (EQUIVALENT transformation)

(20) Obligation:

(21) PisEmptyProof (EQUIVALENT transformation)

(22) TRUE

(23) Obligation:

(24) QDPOrderProof (EQUIVALENT transformation)

(25) Obligation:

(26) QDPOrderProof (EQUIVALENT transformation)

(27) Obligation:

(28) QDPOrderProof (EQUIVALENT transformation)

(29) Obligation:

(30) QDPOrderProof (EQUIVALENT transformation)

(31) Obligation:

(32) PisEmptyProof (EQUIVALENT transformation)

(33) TRUE

(34) Obligation:

(35) QDPOrderProof (EQUIVALENT transformation)

(36) Obligation:

(37) QDPOrderProof (EQUIVALENT transformation)

(38) Obligation:

(39) PisEmptyProof (EQUIVALENT transformation)

(40) TRUE