NO Termination w.r.t. Q proof of /home/cern_httpd/provide/research/cycsrs/tpdb/TPDB-d9b80194f163/SRS_Standard/Zantema_06/01.srs-torpacyc2out-split.srs

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

0 QTRS
↳1 QTRS Reverse (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 7 ms)
        ↳9 QDP
        ↳10 QDPOrderProof (⇔, 78 ms)
        ↳11 QDP
        ↳12 DependencyGraphProof (⇔, 0 ms)
        ↳13 QDP
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QDPSizeChangeProof (⇔, 0 ms)
        ↳18 YES
    ↳19 QDP
        ↳20 UsableRulesProof (⇔, 0 ms)
        ↳21 QDP
        ↳22 QDPSizeChangeProof (⇔, 0 ms)
        ↳23 YES
    ↳24 QDP
        ↳25 UsableRulesProof (⇔, 5 ms)
        ↳26 QDP
        ↳27 NonTerminationLoopProof (⇒, 58 ms)
        ↳28 NO

(0) Obligation:

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

Begin(b(x)) → Wait(Right1(x))
Begin(a(x)) → Wait(Right2(x))
Begin(a(x)) → Wait(Right3(x))
Right1(a(End(x))) → Left(b(r(End(x))))
Right2(r(End(x))) → Left(d(r(End(x))))
Right3(d(End(x))) → Left(a(a(d(End(x)))))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right1(r(x)) → Ar(Right1(x))
Right2(r(x)) → Ar(Right2(x))
Right3(r(x)) → Ar(Right3(x))
Right1(d(x)) → Ad(Right1(x))
Right2(d(x)) → Ad(Right2(x))
Right3(d(x)) → Ad(Right3(x))
Aa(Left(x)) → Left(a(x))
Ab(Left(x)) → Left(b(x))
Ar(Left(x)) → Left(r(x))
Ad(Left(x)) → Left(d(x))
Wait(Left(x)) → Begin(x)
a(b(x)) → b(r(x))
r(a(x)) → d(r(x))
r(x) → d(x)
d(a(x)) → a(a(d(x)))
d(x) → a(x)

Q is empty.

(1) QTRS Reverse (EQUIVALENT transformation)

We applied the QTRS Reverse Processor [REVERSE].

(2) Obligation:

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

b(Begin(x)) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
End(a(Right1(x))) → End(r(b(Left(x))))
End(r(Right2(x))) → End(r(d(Left(x))))
End(d(Right3(x))) → End(d(a(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ar(x)) → r(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Wait(x)) → Begin(x)
b(a(x)) → r(b(x))
a(r(x)) → r(d(x))
r(x) → d(x)
a(d(x)) → d(a(a(x)))
d(x) → a(x)

Q is empty.

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

END(a(Right1(x))) → END(r(b(Left(x))))
END(a(Right1(x))) → R(b(Left(x)))
END(a(Right1(x))) → B(Left(x))
END(a(Right1(x))) → LEFT(x)
END(r(Right2(x))) → END(r(d(Left(x))))
END(r(Right2(x))) → R(d(Left(x)))
END(r(Right2(x))) → D(Left(x))
END(r(Right2(x))) → LEFT(x)
END(d(Right3(x))) → END(d(a(a(Left(x)))))
END(d(Right3(x))) → D(a(a(Left(x))))
END(d(Right3(x))) → A(a(Left(x)))
END(d(Right3(x))) → A(Left(x))
END(d(Right3(x))) → LEFT(x)
LEFT(Aa(x)) → A(Left(x))
LEFT(Aa(x)) → LEFT(x)
LEFT(Ab(x)) → B(Left(x))
LEFT(Ab(x)) → LEFT(x)
LEFT(Ar(x)) → R(Left(x))
LEFT(Ar(x)) → LEFT(x)
LEFT(Ad(x)) → D(Left(x))
LEFT(Ad(x)) → LEFT(x)
B(a(x)) → R(b(x))
B(a(x)) → B(x)
A(r(x)) → R(d(x))
A(r(x)) → D(x)
R(x) → D(x)
A(d(x)) → D(a(a(x)))
A(d(x)) → A(a(x))
A(d(x)) → A(x)
D(x) → A(x)

The TRS R consists of the following rules:

b(Begin(x)) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
End(a(Right1(x))) → End(r(b(Left(x))))
End(r(Right2(x))) → End(r(d(Left(x))))
End(d(Right3(x))) → End(d(a(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ar(x)) → r(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Wait(x)) → Begin(x)
b(a(x)) → r(b(x))
a(r(x)) → r(d(x))
r(x) → d(x)
a(d(x)) → d(a(a(x)))
d(x) → a(x)

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 4 SCCs with 15 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

R(x) → D(x)
D(x) → A(x)
A(r(x)) → R(d(x))
A(r(x)) → D(x)
A(d(x)) → D(a(a(x)))
A(d(x)) → A(a(x))
A(d(x)) → A(x)

The TRS R consists of the following rules:

b(Begin(x)) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
End(a(Right1(x))) → End(r(b(Left(x))))
End(r(Right2(x))) → End(r(d(Left(x))))
End(d(Right3(x))) → End(d(a(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ar(x)) → r(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Wait(x)) → Begin(x)
b(a(x)) → r(b(x))
a(r(x)) → r(d(x))
r(x) → d(x)
a(d(x)) → d(a(a(x)))
d(x) → a(x)

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

(8) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

R(x) → D(x)
D(x) → A(x)
A(r(x)) → R(d(x))
A(r(x)) → D(x)
A(d(x)) → D(a(a(x)))
A(d(x)) → A(a(x))
A(d(x)) → A(x)

The TRS R consists of the following rules:

a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(r(x)) → r(d(x))
r(x) → d(x)
d(x) → a(x)
a(d(x)) → d(a(a(x)))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))

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.


A(r(x)) → R(d(x))
A(r(x)) → D(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(A(x1)) = x1   
POL(Aa(x1)) = x1   
POL(Ad(x1)) = x1   
POL(Ar(x1)) = x1   
POL(Begin(x1)) = x1   
POL(D(x1)) = x1   
POL(R(x1)) = x1   
POL(Right1(x1)) = 0   
POL(Right2(x1)) = 0   
POL(Right3(x1)) = 0   
POL(Wait(x1)) = x1   
POL(a(x1)) = x1   
POL(d(x1)) = x1   
POL(r(x1)) = 1 + x1   

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

a(r(x)) → r(d(x))
r(x) → d(x)
d(x) → a(x)
a(d(x)) → d(a(a(x)))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))

(11) Obligation:

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

R(x) → D(x)
D(x) → A(x)
A(d(x)) → D(a(a(x)))
A(d(x)) → A(a(x))
A(d(x)) → A(x)

The TRS R consists of the following rules:

a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(r(x)) → r(d(x))
r(x) → d(x)
d(x) → a(x)
a(d(x)) → d(a(a(x)))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))

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

(12) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(13) Obligation:

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

A(d(x)) → D(a(a(x)))
D(x) → A(x)
A(d(x)) → A(a(x))
A(d(x)) → A(x)

The TRS R consists of the following rules:

a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(r(x)) → r(d(x))
r(x) → d(x)
d(x) → a(x)
a(d(x)) → d(a(a(x)))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))

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

(14) Obligation:

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

B(a(x)) → B(x)

The TRS R consists of the following rules:

b(Begin(x)) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
End(a(Right1(x))) → End(r(b(Left(x))))
End(r(Right2(x))) → End(r(d(Left(x))))
End(d(Right3(x))) → End(d(a(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ar(x)) → r(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Wait(x)) → Begin(x)
b(a(x)) → r(b(x))
a(r(x)) → r(d(x))
r(x) → d(x)
a(d(x)) → d(a(a(x)))
d(x) → a(x)

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

(15) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

B(a(x)) → B(x)

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

(17) 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:

  • B(a(x)) → B(x)
    The graph contains the following edges 1 > 1

(18) YES

(19) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)
LEFT(Ar(x)) → LEFT(x)
LEFT(Ad(x)) → LEFT(x)

The TRS R consists of the following rules:

b(Begin(x)) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
End(a(Right1(x))) → End(r(b(Left(x))))
End(r(Right2(x))) → End(r(d(Left(x))))
End(d(Right3(x))) → End(d(a(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ar(x)) → r(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Wait(x)) → Begin(x)
b(a(x)) → r(b(x))
a(r(x)) → r(d(x))
r(x) → d(x)
a(d(x)) → d(a(a(x)))
d(x) → a(x)

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

(20) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(21) Obligation:

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

LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → LEFT(x)
LEFT(Ar(x)) → LEFT(x)
LEFT(Ad(x)) → LEFT(x)

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

(22) 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:

  • LEFT(Ab(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

  • LEFT(Aa(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

  • LEFT(Ar(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

  • LEFT(Ad(x)) → LEFT(x)
    The graph contains the following edges 1 > 1

(23) YES

(24) Obligation:

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

END(r(Right2(x))) → END(r(d(Left(x))))
END(a(Right1(x))) → END(r(b(Left(x))))
END(d(Right3(x))) → END(d(a(a(Left(x)))))

The TRS R consists of the following rules:

b(Begin(x)) → Right1(Wait(x))
a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
End(a(Right1(x))) → End(r(b(Left(x))))
End(r(Right2(x))) → End(r(d(Left(x))))
End(d(Right3(x))) → End(d(a(a(Left(x)))))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ar(x)) → r(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Wait(x)) → Begin(x)
b(a(x)) → r(b(x))
a(r(x)) → r(d(x))
r(x) → d(x)
a(d(x)) → d(a(a(x)))
d(x) → a(x)

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

(25) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(26) Obligation:

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

END(r(Right2(x))) → END(r(d(Left(x))))
END(a(Right1(x))) → END(r(b(Left(x))))
END(d(Right3(x))) → END(d(a(a(Left(x)))))

The TRS R consists of the following rules:

Left(Aa(x)) → a(Left(x))
Left(Ab(x)) → b(Left(x))
Left(Ar(x)) → r(Left(x))
Left(Ad(x)) → d(Left(x))
Left(Wait(x)) → Begin(x)
a(Begin(x)) → Right2(Wait(x))
a(Begin(x)) → Right3(Wait(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(r(x)) → r(d(x))
r(x) → d(x)
d(x) → a(x)
a(d(x)) → d(a(a(x)))
d(Right1(x)) → Right1(Ad(x))
d(Right2(x)) → Right2(Ad(x))
d(Right3(x)) → Right3(Ad(x))
r(Right1(x)) → Right1(Ar(x))
r(Right2(x)) → Right2(Ar(x))
r(Right3(x)) → Right3(Ar(x))
b(Begin(x)) → Right1(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(a(x)) → r(b(x))

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

(27) NonTerminationLoopProof (COMPLETE transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = END(r(d(Left(Wait(x))))) evaluates to t =END(r(d(Left(Wait(x)))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

END(r(d(Left(Wait(x))))) → END(r(d(Begin(x))))
with rule Left(Wait(x')) → Begin(x') at position [0,0,0] and matcher [x' / x]

END(r(d(Begin(x)))) → END(r(a(Begin(x))))
with rule d(x') → a(x') at position [0,0] and matcher [x' / Begin(x)]

END(r(a(Begin(x)))) → END(r(Right2(Wait(x))))
with rule a(Begin(x')) → Right2(Wait(x')) at position [0,0] and matcher [x' / x]

END(r(Right2(Wait(x)))) → END(r(d(Left(Wait(x)))))
with rule END(r(Right2(x))) → END(r(d(Left(x))))

Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence


All these steps are and every following step will be a correct step w.r.t to Q.



(28) NO