YES Termination w.r.t. Q proof of Secret_05_TRS_cime1.ari

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

0 QTRS
↳1 DependencyPairsProof (⇔, 13 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 QDPOrderProof (⇔, 126 ms)
↳6 QDP
↳7 DependencyGraphProof (⇔, 0 ms)
↳8 AND
    ↳9 QDP
        ↳10 QDPOrderProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 PisEmptyProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 QDPOrderProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 PisEmptyProof (⇔, 0 ms)
        ↳18 YES

(0) Obligation:

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

sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
sortSu(circ(sortSu(s), sortSu(id))) → sortSu(s)
sortSu(circ(sortSu(id), sortSu(s))) → sortSu(s)
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
te(subst(te(a), sortSu(id))) → te(a)
te(msubst(te(a), sortSu(id))) → te(a)
te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))

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:

SORTSU(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → SORTSU(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
SORTSU(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → TE(msubst(te(a), sortSu(t)))
SORTSU(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → SORTSU(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → SORTSU(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → SORTSU(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
SORTSU(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → SORTSU(circ(sortSu(t), sortSu(u)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → SORTSU(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → SORTSU(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → SORTSU(circ(sortSu(s), sortSu(t)))
TE(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → TE(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
TE(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → SORTSU(circ(sortSu(s), sortSu(t)))

The TRS R consists of the following rules:

sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
sortSu(circ(sortSu(s), sortSu(id))) → sortSu(s)
sortSu(circ(sortSu(id), sortSu(s))) → sortSu(s)
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
te(subst(te(a), sortSu(id))) → te(a)
te(msubst(te(a), sortSu(id))) → te(a)
te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))

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 4 less nodes.

(4) Obligation:

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

SORTSU(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → TE(msubst(te(a), sortSu(t)))
TE(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → TE(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
TE(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → SORTSU(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
SORTSU(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → SORTSU(circ(sortSu(t), sortSu(u)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → SORTSU(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → SORTSU(circ(sortSu(s), sortSu(t)))

The TRS R consists of the following rules:

sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
sortSu(circ(sortSu(s), sortSu(id))) → sortSu(s)
sortSu(circ(sortSu(id), sortSu(s))) → sortSu(s)
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
te(subst(te(a), sortSu(id))) → te(a)
te(msubst(te(a), sortSu(id))) → te(a)
te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))

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.


SORTSU(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → TE(msubst(te(a), sortSu(t)))
TE(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → SORTSU(circ(sortSu(s), sortSu(t)))
SORTSU(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → SORTSU(circ(sortSu(t), sortSu(u)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → SORTSU(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
SORTSU(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → SORTSU(circ(sortSu(s), sortSu(t)))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation with max and min functions [POLO,MAXPOLO]:

POL(SORTSU(x1)) = 1 + x1   
POL(TE(x1)) = x1   
POL(circ(x1, x2)) = x1 + x2   
POL(cons(x1, x2)) = max(x1, x2)   
POL(id) = 0   
POL(lift) = 1   
POL(msubst(x1, x2)) = x1 + x2   
POL(sop(x1)) = 1   
POL(sortSu(x1)) = 1 + x1   
POL(subst(x1, x2)) = x1   
POL(te(x1)) = 1 + x1   

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

te(msubst(te(a), sortSu(id))) → te(a)
te(subst(te(a), sortSu(id))) → te(a)
te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(s), sortSu(id))) → sortSu(s)
sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
sortSu(circ(sortSu(id), sortSu(s))) → sortSu(s)
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))

(6) Obligation:

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

TE(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → TE(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
SORTSU(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → SORTSU(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))

The TRS R consists of the following rules:

sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
sortSu(circ(sortSu(s), sortSu(id))) → sortSu(s)
sortSu(circ(sortSu(id), sortSu(s))) → sortSu(s)
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
te(subst(te(a), sortSu(id))) → te(a)
te(msubst(te(a), sortSu(id))) → te(a)
te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.

(8) Complex Obligation (AND)

(9) Obligation:

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

SORTSU(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → SORTSU(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))

The TRS R consists of the following rules:

sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
sortSu(circ(sortSu(s), sortSu(id))) → sortSu(s)
sortSu(circ(sortSu(id), sortSu(s))) → sortSu(s)
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
te(subst(te(a), sortSu(id))) → te(a)
te(msubst(te(a), sortSu(id))) → te(a)
te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))

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

(10) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


SORTSU(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → SORTSU(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SORTSU(x1)  =  x1
circ(x1, x2)  =  circ(x1, x2)
sortSu(x1)  =  x1
cons(x1, x2)  =  cons
id  =  id

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

circ_2=1
cons=1
id=2

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

sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(s), sortSu(id))) → sortSu(s)
sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
sortSu(circ(sortSu(id), sortSu(s))) → sortSu(s)
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))

(11) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
sortSu(circ(sortSu(s), sortSu(id))) → sortSu(s)
sortSu(circ(sortSu(id), sortSu(s))) → sortSu(s)
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
te(subst(te(a), sortSu(id))) → te(a)
te(msubst(te(a), sortSu(id))) → te(a)
te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))

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

(12) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(13) YES

(14) Obligation:

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

TE(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → TE(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))

The TRS R consists of the following rules:

sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
sortSu(circ(sortSu(s), sortSu(id))) → sortSu(s)
sortSu(circ(sortSu(id), sortSu(s))) → sortSu(s)
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
te(subst(te(a), sortSu(id))) → te(a)
te(msubst(te(a), sortSu(id))) → te(a)
te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))

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

(15) QDPOrderProof (EQUIVALENT transformation)

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


The following pairs can be oriented strictly and are deleted.


TE(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → TE(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
TE(x1)  =  x1
msubst(x1, x2)  =  x1
te(x1)  =  te(x1)
subst(x1, x2)  =  x1
sortSu(x1)  =  sortSu
id  =  id

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

dummyConstant=1
te_1=1

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

te(msubst(te(a), sortSu(id))) → te(a)
te(subst(te(a), sortSu(id))) → te(a)
te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))

(16) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sortSu(circ(sortSu(cons(te(a), sortSu(s))), sortSu(t))) → sortSu(cons(te(msubst(te(a), sortSu(t))), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(te(a), sortSu(t))))) → sortSu(cons(te(a), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(cons(sop(lift), sortSu(t))))) → sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t)))))
sortSu(circ(sortSu(circ(sortSu(s), sortSu(t))), sortSu(u))) → sortSu(circ(sortSu(s), sortSu(circ(sortSu(t), sortSu(u)))))
sortSu(circ(sortSu(s), sortSu(id))) → sortSu(s)
sortSu(circ(sortSu(id), sortSu(s))) → sortSu(s)
sortSu(circ(sortSu(cons(sop(lift), sortSu(s))), sortSu(circ(sortSu(cons(sop(lift), sortSu(t))), sortSu(u))))) → sortSu(circ(sortSu(cons(sop(lift), sortSu(circ(sortSu(s), sortSu(t))))), sortSu(u)))
te(subst(te(a), sortSu(id))) → te(a)
te(msubst(te(a), sortSu(id))) → te(a)
te(msubst(te(msubst(te(a), sortSu(s))), sortSu(t))) → te(msubst(te(a), sortSu(circ(sortSu(s), sortSu(t)))))

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

(17) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(18) YES