(0) Obligation:

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

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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(U21(tt, M, N)) → S(plus(N, M))
ACTIVE(U21(tt, M, N)) → PLUS(N, M)
ACTIVE(isNat(plus(V1, V2))) → AND(isNat(V1), isNat(V2))
ACTIVE(isNat(plus(V1, V2))) → ISNAT(V1)
ACTIVE(isNat(plus(V1, V2))) → ISNAT(V2)
ACTIVE(isNat(s(V1))) → ISNAT(V1)
ACTIVE(plus(N, 0)) → U111(isNat(N), N)
ACTIVE(plus(N, 0)) → ISNAT(N)
ACTIVE(plus(N, s(M))) → U211(and(isNat(M), isNat(N)), M, N)
ACTIVE(plus(N, s(M))) → AND(isNat(M), isNat(N))
ACTIVE(plus(N, s(M))) → ISNAT(M)
ACTIVE(plus(N, s(M))) → ISNAT(N)
ACTIVE(U11(X1, X2)) → U111(active(X1), X2)
ACTIVE(U11(X1, X2)) → ACTIVE(X1)
ACTIVE(U21(X1, X2, X3)) → U211(active(X1), X2, X3)
ACTIVE(U21(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)
ACTIVE(and(X1, X2)) → AND(active(X1), X2)
ACTIVE(and(X1, X2)) → ACTIVE(X1)
U111(mark(X1), X2) → U111(X1, X2)
U211(mark(X1), X2, X3) → U211(X1, X2, X3)
S(mark(X)) → S(X)
PLUS(mark(X1), X2) → PLUS(X1, X2)
PLUS(X1, mark(X2)) → PLUS(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
PROPER(U11(X1, X2)) → U111(proper(X1), proper(X2))
PROPER(U11(X1, X2)) → PROPER(X1)
PROPER(U11(X1, X2)) → PROPER(X2)
PROPER(U21(X1, X2, X3)) → U211(proper(X1), proper(X2), proper(X3))
PROPER(U21(X1, X2, X3)) → PROPER(X1)
PROPER(U21(X1, X2, X3)) → PROPER(X2)
PROPER(U21(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)
PROPER(and(X1, X2)) → AND(proper(X1), proper(X2))
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(isNat(X)) → ISNAT(proper(X))
PROPER(isNat(X)) → PROPER(X)
U111(ok(X1), ok(X2)) → U111(X1, X2)
U211(ok(X1), ok(X2), ok(X3)) → U211(X1, X2, X3)
S(ok(X)) → S(X)
PLUS(ok(X1), ok(X2)) → PLUS(X1, X2)
AND(ok(X1), ok(X2)) → AND(X1, X2)
ISNAT(ok(X)) → ISNAT(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(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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 9 SCCs with 26 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

ISNAT(ok(X)) → ISNAT(X)

The TRS R consists of the following rules:

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.


ISNAT(ok(X)) → ISNAT(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
ok1 > ISNAT1

Status:
ISNAT1: [1]
ok1: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(11) 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)
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)
ok(x1)  =  ok(x1)
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
AND1: [1]
ok1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(16) 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

(17) 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, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(18) 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

(19) Obligation:

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

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE

(22) 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, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(23) 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

(24) 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, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(25) 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

(26) Obligation:

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

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(27) PisEmptyProof (EQUIVALENT transformation)

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

(28) TRUE

(29) Obligation:

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

U211(ok(X1), ok(X2), ok(X3)) → U211(X1, X2, X3)
U211(mark(X1), X2, X3) → U211(X1, X2, X3)

The TRS R consists of the following rules:

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(30) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


U211(ok(X1), ok(X2), ok(X3)) → U211(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U211(x1, x2, x3)  =  U211(x3)
ok(x1)  =  ok(x1)
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Precedence:
mark1 > U21^11

Status:
U21^11: [1]
ok1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(31) Obligation:

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

U211(mark(X1), X2, X3) → U211(X1, X2, X3)

The TRS R consists of the following rules:

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(32) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


U211(mark(X1), X2, X3) → U211(X1, X2, X3)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
mark1 > U21^13

Status:
U21^13: [3,1,2]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(33) Obligation:

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

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(34) PisEmptyProof (EQUIVALENT transformation)

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

(35) TRUE

(36) Obligation:

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

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

The TRS R consists of the following rules:

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(37) 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)) → U111(X1, X2)
U111(mark(X1), X2) → U111(X1, X2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
U111(x1, x2)  =  U111(x1)
ok(x1)  =  ok(x1)
mark(x1)  =  mark(x1)

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
U11^11: [1]
ok1: [1]
mark1: [1]

The following usable rules [FROCOS05] were oriented: none

(38) Obligation:

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

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(39) PisEmptyProof (EQUIVALENT transformation)

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

(40) TRUE

(41) Obligation:

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

PROPER(U11(X1, X2)) → PROPER(X2)
PROPER(U11(X1, X2)) → PROPER(X1)
PROPER(U21(X1, X2, X3)) → PROPER(X1)
PROPER(U21(X1, X2, X3)) → PROPER(X2)
PROPER(U21(X1, X2, X3)) → PROPER(X3)
PROPER(s(X)) → PROPER(X)
PROPER(plus(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(X1, X2)) → PROPER(X2)
PROPER(isNat(X)) → PROPER(X)

The TRS R consists of the following rules:

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(42) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(U11(X1, X2)) → PROPER(X2)
PROPER(U11(X1, X2)) → PROPER(X1)
PROPER(U21(X1, X2, X3)) → PROPER(X1)
PROPER(U21(X1, X2, X3)) → PROPER(X2)
PROPER(U21(X1, X2, X3)) → PROPER(X3)
PROPER(s(X)) → PROPER(X)
PROPER(plus(X1, X2)) → PROPER(X1)
PROPER(plus(X1, X2)) → PROPER(X2)
PROPER(and(X1, X2)) → PROPER(X1)
PROPER(and(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)  =  U11(x1, x2)
U21(x1, x2, x3)  =  U21(x1, x2, x3)
s(x1)  =  s(x1)
plus(x1, x2)  =  plus(x1, x2)
and(x1, x2)  =  and(x1, x2)
isNat(x1)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
U112: [1,2]
U213: [1,2,3]
s1: [1]
plus2: [1,2]
and2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(43) Obligation:

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

PROPER(isNat(X)) → PROPER(X)

The TRS R consists of the following rules:

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(44) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PROPER(isNat(X)) → PROPER(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
isNat1 > PROPER1

Status:
PROPER1: [1]
isNat1: [1]

The following usable rules [FROCOS05] were oriented: none

(45) Obligation:

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

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(46) PisEmptyProof (EQUIVALENT transformation)

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

(47) TRUE

(48) Obligation:

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

ACTIVE(U21(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(U11(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(plus(X1, X2)) → ACTIVE(X1)
ACTIVE(plus(X1, X2)) → ACTIVE(X2)
ACTIVE(and(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(U21(X1, X2, X3)) → ACTIVE(X1)
ACTIVE(U11(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
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
U21(x1, x2, x3)  =  U21(x1)
U11(x1, x2)  =  U11(x1)
s(x1)  =  s(x1)
plus(x1, x2)  =  plus(x1, x2)
and(x1, x2)  =  x1

Lexicographic path order with status [LPO].
Precedence:
trivial

Status:
U211: [1]
U111: [1]
s1: [1]
plus2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(50) Obligation:

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

ACTIVE(and(X1, X2)) → ACTIVE(X1)

The TRS R consists of the following rules:

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACTIVE(and(X1, X2)) → ACTIVE(X1)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic path order with status [LPO].
Precedence:
and2 > ACTIVE1

Status:
ACTIVE1: [1]
and2: [1,2]

The following usable rules [FROCOS05] were oriented: none

(52) Obligation:

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

active(U11(tt, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.

(53) PisEmptyProof (EQUIVALENT transformation)

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

(54) TRUE

(55) 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, N)) → mark(N)
active(U21(tt, M, N)) → mark(s(plus(N, M)))
active(and(tt, X)) → mark(X)
active(isNat(0)) → mark(tt)
active(isNat(plus(V1, V2))) → mark(and(isNat(V1), isNat(V2)))
active(isNat(s(V1))) → mark(isNat(V1))
active(plus(N, 0)) → mark(U11(isNat(N), N))
active(plus(N, s(M))) → mark(U21(and(isNat(M), isNat(N)), M, N))
active(U11(X1, X2)) → U11(active(X1), X2)
active(U21(X1, X2, X3)) → U21(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))
active(and(X1, X2)) → and(active(X1), X2)
U11(mark(X1), X2) → mark(U11(X1, X2))
U21(mark(X1), X2, X3) → mark(U21(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))
and(mark(X1), X2) → mark(and(X1, X2))
proper(U11(X1, X2)) → U11(proper(X1), proper(X2))
proper(tt) → ok(tt)
proper(U21(X1, X2, X3)) → U21(proper(X1), proper(X2), proper(X3))
proper(s(X)) → s(proper(X))
proper(plus(X1, X2)) → plus(proper(X1), proper(X2))
proper(and(X1, X2)) → and(proper(X1), proper(X2))
proper(isNat(X)) → isNat(proper(X))
proper(0) → ok(0)
U11(ok(X1), ok(X2)) → ok(U11(X1, X2))
U21(ok(X1), ok(X2), ok(X3)) → ok(U21(X1, X2, X3))
s(ok(X)) → ok(s(X))
plus(ok(X1), ok(X2)) → ok(plus(X1, X2))
and(ok(X1), ok(X2)) → ok(and(X1, X2))
isNat(ok(X)) → ok(isNat(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.