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, M, N) → a__U12(tt, M, N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2, X3) → U11(X1, X2, X3)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)

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, M, N) → A__U12(tt, M, N)
A__U12(tt, M, N) → A__PLUS(mark(N), mark(M))
A__U12(tt, M, N) → MARK(N)
A__U12(tt, M, N) → MARK(M)
A__PLUS(N, 0) → MARK(N)
A__PLUS(N, s(M)) → A__U11(tt, M, N)
MARK(U11(X1, X2, X3)) → A__U11(mark(X1), X2, X3)
MARK(U11(X1, X2, X3)) → MARK(X1)
MARK(U12(X1, X2, X3)) → A__U12(mark(X1), X2, X3)
MARK(U12(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(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__U11(tt, M, N) → a__U12(tt, M, N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2, X3) → U11(X1, X2, X3)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)

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__U12(tt, M, N) → MARK(N)
A__U12(tt, M, N) → MARK(M)
A__PLUS(N, 0) → MARK(N)
A__PLUS(N, s(M)) → A__U11(tt, M, N)
MARK(U11(X1, X2, X3)) → A__U11(mark(X1), X2, X3)
MARK(U11(X1, X2, X3)) → MARK(X1)
MARK(U12(X1, X2, X3)) → A__U12(mark(X1), X2, X3)
MARK(U12(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(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, x3)  =  A__U11(x2, x3)
tt  =  tt
A__U12(x1, x2, x3)  =  A__U12(x2, x3)
A__PLUS(x1, x2)  =  A__PLUS(x1, x2)
mark(x1)  =  x1
MARK(x1)  =  MARK(x1)
0  =  0
s(x1)  =  s(x1)
U11(x1, x2, x3)  =  U11(x1, x2, x3)
U12(x1, x2, x3)  =  U12(x1, x2, x3)
plus(x1, x2)  =  plus(x1, x2)
a__U11(x1, x2, x3)  =  a__U11(x1, x2, x3)
a__plus(x1, x2)  =  a__plus(x1, x2)
a__U12(x1, x2, x3)  =  a__U12(x1, x2, x3)

Recursive Path Order [RPO].
Precedence:
[AU112, AU122, APLUS2, MARK1, 0] > [tt, s1]
[U113, U123, plus2, aU113, aplus2, aU123] > [tt, s1]


The following usable rules [FROCOS05] were oriented:

a__U11(X1, X2, X3) → U11(X1, X2, X3)
mark(0) → 0
a__plus(X1, X2) → plus(X1, X2)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
a__U11(tt, M, N) → a__U12(tt, M, N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
a__plus(N, 0) → mark(N)
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)

(4) Obligation:

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

A__U11(tt, M, N) → A__U12(tt, M, N)
A__U12(tt, M, N) → A__PLUS(mark(N), mark(M))

The TRS R consists of the following rules:

a__U11(tt, M, N) → a__U12(tt, M, N)
a__U12(tt, M, N) → s(a__plus(mark(N), mark(M)))
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → a__U11(tt, M, N)
mark(U11(X1, X2, X3)) → a__U11(mark(X1), X2, X3)
mark(U12(X1, X2, X3)) → a__U12(mark(X1), X2, X3)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(tt) → tt
mark(s(X)) → s(mark(X))
mark(0) → 0
a__U11(X1, X2, X3) → U11(X1, X2, X3)
a__U12(X1, X2, X3) → U12(X1, X2, X3)
a__plus(X1, X2) → plus(X1, X2)

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 2 less nodes.

(6) TRUE