Termination w.r.t. Q proof of /tmp/tmpYjtUkM/MYNAT_nosorts

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__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
a__x(N, 0) → 0
a__x(N, s(M)) → a__plus(a__x(mark(N), mark(M)), mark(N))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(x(X1, X2)) → a__x(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)
a__x(X1, X2) → x(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__AND(tt, X) → MARK(X)
A__PLUS(N, 0) → MARK(N)
A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))
A__PLUS(N, s(M)) → MARK(N)
A__PLUS(N, s(M)) → MARK(M)
A__X(N, s(M)) → A__PLUS(a__x(mark(N), mark(M)), mark(N))
A__X(N, s(M)) → A__X(mark(N), mark(M))
A__X(N, s(M)) → MARK(N)
A__X(N, s(M)) → MARK(M)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → 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(x(X1, X2)) → A__X(mark(X1), mark(X2))
MARK(x(X1, X2)) → MARK(X1)
MARK(x(X1, X2)) → MARK(X2)
MARK(s(X)) → MARK(X)

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
a__x(N, 0) → 0
a__x(N, s(M)) → a__plus(a__x(mark(N), mark(M)), mark(N))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(x(X1, X2)) → a__x(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)
a__x(X1, X2) → x(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__AND(tt, X) → MARK(X)
A__PLUS(N, 0) → MARK(N)
A__PLUS(N, s(M)) → A__PLUS(mark(N), mark(M))
A__PLUS(N, s(M)) → MARK(N)
A__PLUS(N, s(M)) → MARK(M)
A__X(N, s(M)) → A__PLUS(a__x(mark(N), mark(M)), mark(N))
A__X(N, s(M)) → A__X(mark(N), mark(M))
A__X(N, s(M)) → MARK(N)
A__X(N, s(M)) → MARK(M)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → 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(x(X1, X2)) → MARK(X1)
MARK(x(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__AND(x1, x2) = A__AND(x1, x2)
tt = tt
MARK(x1) = x1
A__PLUS(x1, x2) = A__PLUS(x1, x2)
0 = 0
s(x1) = s(x1)
mark(x1) = x1
A__X(x1, x2) = A__X(x1, x2)
a__x(x1, x2) = a__x(x1, x2)
and(x1, x2) = and(x1, x2)
plus(x1, x2) = plus(x1, x2)
x(x1, x2) = x(x1, x2)
a__and(x1, x2) = a__and(x1, x2)
a__plus(x1, x2) = a__plus(x1, x2)

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

tt > [A_{_AND₂}, A_{_PLUS₂}, 0]
[A_{_X₂}, a_{_x₂}, x₂] > [plus₂, a_{_plus₂}] > s₁ > [A_{_AND₂}, A_{_PLUS₂}, 0]
[and₂, a_{_and₂}] > [A_{_AND₂}, A_{_PLUS₂}, 0]

Status:

A_{_AND₂}: [1,2]
tt: []
A_{_PLUS₂}: [2,1]
0: []
s₁: [1]
A_{_X₂}: [2,1]
a_{_x₂}: [2,1]
and₂: [2,1]
plus₂: [2,1]
x₂: [2,1]
a_{_and₂}: [2,1]
a_{_plus₂}: [2,1]

The following usable rules [FROCOS05] were oriented:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
a__x(N, 0) → 0
a__x(N, s(M)) → a__plus(a__x(mark(N), mark(M)), mark(N))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(x(X1, X2)) → a__x(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)
a__x(X1, X2) → x(X1, X2)

(4) Obligation:

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

MARK(x(X1, X2)) → A__X(mark(X1), mark(X2))

The TRS R consists of the following rules:

a__and(tt, X) → mark(X)
a__plus(N, 0) → mark(N)
a__plus(N, s(M)) → s(a__plus(mark(N), mark(M)))
a__x(N, 0) → 0
a__x(N, s(M)) → a__plus(a__x(mark(N), mark(M)), mark(N))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(plus(X1, X2)) → a__plus(mark(X1), mark(X2))
mark(x(X1, X2)) → a__x(mark(X1), mark(X2))
mark(tt) → tt
mark(0) → 0
mark(s(X)) → s(mark(X))
a__and(X1, X2) → and(X1, X2)
a__plus(X1, X2) → plus(X1, X2)
a__x(X1, X2) → x(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 1 less node.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) QDPOrderProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) TRUE