YES Termination w.r.t. Q proof of Mixed_TRS_sigma.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 486 ms)
↳6 QDP
↳7 QDPSizeChangeProof (⇔, 0 ms)
↳8 YES

(0) Obligation:

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

comp(s, id) → s
cons(one, shift) → id
cons(comp(one, s), comp(shift, s)) → s
comp(one, cons(s, t)) → s
comp(shift, cons(s, t)) → t
comp(abs(s), t) → abs(comp(s, cons(one, comp(t, shift))))
comp(cons(s, t), u) → cons(comp(s, u), comp(t, u))
comp(id, s) → s
comp(comp(s, t), u) → comp(s, comp(t, u))

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:

COMP(abs(s), t) → COMP(s, cons(one, comp(t, shift)))
COMP(abs(s), t) → CONS(one, comp(t, shift))
COMP(abs(s), t) → COMP(t, shift)
COMP(cons(s, t), u) → CONS(comp(s, u), comp(t, u))
COMP(cons(s, t), u) → COMP(s, u)
COMP(cons(s, t), u) → COMP(t, u)
COMP(comp(s, t), u) → COMP(s, comp(t, u))
COMP(comp(s, t), u) → COMP(t, u)

The TRS R consists of the following rules:

comp(s, id) → s
cons(one, shift) → id
cons(comp(one, s), comp(shift, s)) → s
comp(one, cons(s, t)) → s
comp(shift, cons(s, t)) → t
comp(abs(s), t) → abs(comp(s, cons(one, comp(t, shift))))
comp(cons(s, t), u) → cons(comp(s, u), comp(t, u))
comp(id, s) → s
comp(comp(s, t), u) → comp(s, comp(t, u))

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 1 SCC with 2 less nodes.

(4) Obligation:

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

COMP(abs(s), t) → COMP(t, shift)
COMP(abs(s), t) → COMP(s, cons(one, comp(t, shift)))
COMP(cons(s, t), u) → COMP(s, u)
COMP(cons(s, t), u) → COMP(t, u)
COMP(comp(s, t), u) → COMP(s, comp(t, u))
COMP(comp(s, t), u) → COMP(t, u)

The TRS R consists of the following rules:

comp(s, id) → s
cons(one, shift) → id
cons(comp(one, s), comp(shift, s)) → s
comp(one, cons(s, t)) → s
comp(shift, cons(s, t)) → t
comp(abs(s), t) → abs(comp(s, cons(one, comp(t, shift))))
comp(cons(s, t), u) → cons(comp(s, u), comp(t, u))
comp(id, s) → s
comp(comp(s, t), u) → comp(s, comp(t, u))

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


COMP(abs(s), t) → COMP(t, shift)
COMP(abs(s), t) → COMP(s, cons(one, comp(t, shift)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation with max and min functions [POLO,MAXPOLO]:

POL(COMP(x1, x2)) = x1 + x2   
POL(abs(x1)) = 1 + x1   
POL(comp(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = max(x1, x2)   
POL(id) = 0   
POL(one) = 0   
POL(shift) = 0   

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

comp(abs(s), t) → abs(comp(s, cons(one, comp(t, shift))))
comp(cons(s, t), u) → cons(comp(s, u), comp(t, u))
comp(id, s) → s
comp(comp(s, t), u) → comp(s, comp(t, u))
cons(one, shift) → id
comp(s, id) → s
comp(one, cons(s, t)) → s
comp(shift, cons(s, t)) → t
cons(comp(one, s), comp(shift, s)) → s

(6) Obligation:

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

COMP(cons(s, t), u) → COMP(s, u)
COMP(cons(s, t), u) → COMP(t, u)
COMP(comp(s, t), u) → COMP(s, comp(t, u))
COMP(comp(s, t), u) → COMP(t, u)

The TRS R consists of the following rules:

comp(s, id) → s
cons(one, shift) → id
cons(comp(one, s), comp(shift, s)) → s
comp(one, cons(s, t)) → s
comp(shift, cons(s, t)) → t
comp(abs(s), t) → abs(comp(s, cons(one, comp(t, shift))))
comp(cons(s, t), u) → cons(comp(s, u), comp(t, u))
comp(id, s) → s
comp(comp(s, t), u) → comp(s, comp(t, u))

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

(7) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

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

  • COMP(cons(s, t), u) → COMP(s, u)
    The graph contains the following edges 1 > 1, 2 >= 2

  • COMP(cons(s, t), u) → COMP(t, u)
    The graph contains the following edges 1 > 1, 2 >= 2

  • COMP(comp(s, t), u) → COMP(s, comp(t, u))
    The graph contains the following edges 1 > 1

  • COMP(comp(s, t), u) → COMP(t, u)
    The graph contains the following edges 1 > 1, 2 >= 2

(8) YES