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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 TRUE

(0) Obligation:

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

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(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:

A__U11(tt, N) → MARK(N)
A__U21(tt, M, N) → A__PLUS(mark(N), mark(M))
A__U21(tt, M, N) → MARK(N)
A__U21(tt, M, N) → MARK(M)
A__AND(tt, X) → MARK(X)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__PLUS(N, 0) → A__U11(a__isNat(N), N)
A__PLUS(N, 0) → A__ISNAT(N)
A__PLUS(N, s(M)) → A__U21(a__and(a__isNat(M), isNat(N)), M, N)
A__PLUS(N, s(M)) → A__AND(a__isNat(M), isNat(N))
A__PLUS(N, s(M)) → A__ISNAT(M)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U11(X1, X2)) → MARK(X1)
MARK(U21(X1, X2, X3)) → A__U21(mark(X1), X2, X3)
MARK(U21(X1, X2, X3)) → MARK(X1)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(isNat(X)) → A__ISNAT(X)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


A__U11(tt, N) → MARK(N)
A__U21(tt, M, N) → MARK(N)
A__U21(tt, M, N) → MARK(M)
A__AND(tt, X) → MARK(X)
A__ISNAT(plus(V1, V2)) → A__AND(a__isNat(V1), isNat(V2))
A__ISNAT(plus(V1, V2)) → A__ISNAT(V1)
A__ISNAT(s(V1)) → A__ISNAT(V1)
A__PLUS(N, 0) → A__U11(a__isNat(N), N)
A__PLUS(N, 0) → A__ISNAT(N)
A__PLUS(N, s(M)) → A__U21(a__and(a__isNat(M), isNat(N)), M, N)
A__PLUS(N, s(M)) → A__AND(a__isNat(M), isNat(N))
A__PLUS(N, s(M)) → A__ISNAT(M)
MARK(U11(X1, X2)) → A__U11(mark(X1), X2)
MARK(U11(X1, X2)) → MARK(X1)
MARK(U21(X1, X2, X3)) → A__U21(mark(X1), X2, X3)
MARK(U21(X1, X2, X3)) → MARK(X1)
MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2))
MARK(plus(X1, X2)) → MARK(X1)
MARK(plus(X1, X2)) → MARK(X2)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(isNat(X)) → A__ISNAT(X)
MARK(s(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
A__U11(x1, x2)  =  A__U11(x1, x2)
tt  =  tt
MARK(x1)  =  MARK(x1)
A__U21(x1, x2, x3)  =  A__U21(x2, x3)
A__PLUS(x1, x2)  =  A__PLUS(x1, x2)
mark(x1)  =  x1
A__AND(x1, x2)  =  A__AND(x1, x2)
A__ISNAT(x1)  =  A__ISNAT(x1)
plus(x1, x2)  =  plus(x1, x2)
a__isNat(x1)  =  a__isNat(x1)
isNat(x1)  =  isNat(x1)
s(x1)  =  s(x1)
0  =  0
a__and(x1, x2)  =  a__and(x1, x2)
U11(x1, x2)  =  U11(x1, x2)
U21(x1, x2, x3)  =  U21(x1, x2, x3)
and(x1, x2)  =  and(x1, x2)
a__U11(x1, x2)  =  a__U11(x1, x2)
a__U21(x1, x2, x3)  =  a__U21(x1, x2, x3)
a__plus(x1, x2)  =  a__plus(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[plus2, U213, aU213, aplus2] > s1 > [aisNat1, isNat1] > tt > [0, U112, aU112]
[plus2, U213, aU213, aplus2] > [aand2, and2] > [AU212, APLUS2, AAND2] > [AU112, MARK1, AISNAT1] > [aisNat1, isNat1] > tt > [0, U112, aU112]

Status:
plus2: [2,1]
APLUS2: [2,1]
AU112: [2,1]
AISNAT1: [1]
aand2: [2,1]
aU112: [2,1]
aU213: [2,3,1]
U112: [2,1]
and2: [2,1]
0: []
aplus2: [2,1]
AAND2: [1,2]
MARK1: [1]
AU212: [1,2]
tt: []
aisNat1: [1]
s1: [1]
isNat1: [1]
U213: [2,3,1]


The following usable rules [FROCOS05] were oriented:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

(4) Obligation:

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

A__U21(tt, M, N) → A__PLUS(mark(N), mark(M))

The TRS R consists of the following rules:

a__U11(tt, N) → mark(N)
a__U21(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__and(tt, X) → mark(X)
a__isNat(0) → tt
a__isNat(plus(V1, V2)) → a__and(a__isNat(V1), isNat(V2))
a__isNat(s(V1)) → a__isNat(V1)
a__plus(N, 0) → a__U11(a__isNat(N), N)
a__plus(N, s(M)) → a__U21(a__and(a__isNat(M), isNat(N)), M, N)
mark(U11(X1, X2)) → a__U11(mark(X1), X2)
mark(U21(X1, X2, X3)) → a__U21(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(isNat(X)) → a__isNat(X)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2) → U11(X1, X2)
a__U21(X1, X2, X3) → U21(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__isNat(X) → isNat(X)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(6) TRUE