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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPSizeChangeProof (⇔)
↳4 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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Combined order from the following AFS and order.
mark(x1)  =  x1
U11(x1, x2, x3)  =  U11(x1, x2, x3)
a__U11(x1, x2, x3)  =  a__U11(x1, x2, x3)
U12(x1, x2, x3)  =  U12(x1, x2, x3)
a__U12(x1, x2, x3)  =  a__U12(x1, x2, x3)
a__plus(x1, x2)  =  a__plus(x1, x2)
0  =  0
plus(x1, x2)  =  plus(x1, x2)
tt  =  tt
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

[U113, aU113, U123, aU123, aplus2, plus2, tt] > s1

Status:
U113: [3,2,1]
aU113: [3,2,1]
U123: [3,2,1]
aU123: [3,2,1]
aplus2: [1,2]
0: []
plus2: [1,2]
tt: []
s1: [1]

AFS:
mark(x1)  =  x1
U11(x1, x2, x3)  =  U11(x1, x2, x3)
a__U11(x1, x2, x3)  =  a__U11(x1, x2, x3)
U12(x1, x2, x3)  =  U12(x1, x2, x3)
a__U12(x1, x2, x3)  =  a__U12(x1, x2, x3)
a__plus(x1, x2)  =  a__plus(x1, x2)
0  =  0
plus(x1, x2)  =  plus(x1, x2)
tt  =  tt
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • A__PLUS(N, s(M)) → A__U11(tt, M, N) (allowed arguments on rhs = {2, 3})
    The graph contains the following edges 2 > 2, 1 >= 3

  • MARK(U11(X1, X2, X3)) → A__U11(mark(X1), X2, X3) (allowed arguments on rhs = {2, 3})
    The graph contains the following edges 1 > 2, 1 > 3

  • A__PLUS(N, 0) → MARK(N) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 >= 1

  • MARK(U12(X1, X2, X3)) → A__U12(mark(X1), X2, X3) (allowed arguments on rhs = {2, 3})
    The graph contains the following edges 1 > 2, 1 > 3

  • A__U11(tt, M, N) → A__U12(tt, M, N) (allowed arguments on rhs = {1, 2, 3})
    The graph contains the following edges 1 >= 1, 2 >= 2, 3 >= 3

  • A__U12(tt, M, N) → A__PLUS(mark(N), mark(M)) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 3 >= 1, 2 >= 2

  • MARK(plus(X1, X2)) → A__PLUS(mark(X1), mark(X2)) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 1 > 2

  • A__U12(tt, M, N) → MARK(N) (allowed arguments on rhs = {1})
    The graph contains the following edges 3 >= 1

  • A__U12(tt, M, N) → MARK(M) (allowed arguments on rhs = {1})
    The graph contains the following edges 2 >= 1

  • MARK(U11(X1, X2, X3)) → MARK(X1) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • MARK(U12(X1, X2, X3)) → MARK(X1) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • MARK(plus(X1, X2)) → MARK(X1) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • MARK(plus(X1, X2)) → MARK(X2) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • MARK(s(X)) → MARK(X) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05].


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

(4) TRUE