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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 UsableRulesReductionPairsProof (⇔)
    ↳11 QDP
    ↳12 DependencyGraphProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 UsableRulesReductionPairsProof (⇔)
    ↳18 QDP
    ↳19 UsableRulesProof (⇔)
    ↳20 QDP
    ↳21 Narrowing (⇔)
    ↳22 QDP
    ↳23 Narrowing (⇔)
    ↳24 QDP
    ↳25 DependencyGraphProof (⇔)
    ↳26 QDP
    ↳27 Narrowing (⇔)
    ↳28 QDP
    ↳29 DependencyGraphProof (⇔)
    ↳30 QDP
    ↳31 NonTerminationProof (⇔)
    ↳32 FALSE

(0) Obligation:

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

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
take(0, IL) → nil
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(U11(x1, x2)) = x1 + x2   
POL(U12(x1, x2)) = x1 + x2   
POL(U21(x1, x2, x3, x4)) = x1 + x2 + x3 + x4   
POL(U22(x1, x2, x3, x4)) = x1 + x2 + x3 + x4   
POL(U23(x1, x2, x3, x4)) = x1 + x2 + x3 + x4   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(length(x1)) = 1 + x1   
POL(n__take(x1, x2)) = 1 + x1 + x2   
POL(n__zeros) = 0   
POL(nil) = 0   
POL(s(x1)) = x1   
POL(take(x1, x2)) = 1 + x1 + x2   
POL(tt) = 1   
POL(zeros) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

length(nil) → 0
take(0, IL) → nil


(2) Obligation:

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

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(cons(N, L)) → U11(tt, activate(L))
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → 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:

U111(tt, L) → U121(tt, activate(L))
U111(tt, L) → ACTIVATE(L)
U121(tt, L) → LENGTH(activate(L))
U121(tt, L) → ACTIVATE(L)
U211(tt, IL, M, N) → U221(tt, activate(IL), activate(M), activate(N))
U211(tt, IL, M, N) → ACTIVATE(IL)
U211(tt, IL, M, N) → ACTIVATE(M)
U211(tt, IL, M, N) → ACTIVATE(N)
U221(tt, IL, M, N) → U231(tt, activate(IL), activate(M), activate(N))
U221(tt, IL, M, N) → ACTIVATE(IL)
U221(tt, IL, M, N) → ACTIVATE(M)
U221(tt, IL, M, N) → ACTIVATE(N)
U231(tt, IL, M, N) → ACTIVATE(N)
U231(tt, IL, M, N) → ACTIVATE(M)
U231(tt, IL, M, N) → ACTIVATE(IL)
LENGTH(cons(N, L)) → U111(tt, activate(L))
LENGTH(cons(N, L)) → ACTIVATE(L)
TAKE(s(M), cons(N, IL)) → U211(tt, activate(IL), M, N)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
ACTIVATE(n__zeros) → ZEROS
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(cons(N, L)) → U11(tt, activate(L))
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → 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 2 SCCs with 4 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

U221(tt, IL, M, N) → U231(tt, activate(IL), activate(M), activate(N))
U231(tt, IL, M, N) → ACTIVATE(N)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(M), cons(N, IL)) → U211(tt, activate(IL), M, N)
U211(tt, IL, M, N) → U221(tt, activate(IL), activate(M), activate(N))
U221(tt, IL, M, N) → ACTIVATE(IL)
U221(tt, IL, M, N) → ACTIVATE(M)
U221(tt, IL, M, N) → ACTIVATE(N)
U211(tt, IL, M, N) → ACTIVATE(IL)
U211(tt, IL, M, N) → ACTIVATE(M)
U211(tt, IL, M, N) → ACTIVATE(N)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
U231(tt, IL, M, N) → ACTIVATE(M)
U231(tt, IL, M, N) → ACTIVATE(IL)

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(cons(N, L)) → U11(tt, activate(L))
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → 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:

U221(tt, IL, M, N) → U231(tt, activate(IL), activate(M), activate(N))
U231(tt, IL, M, N) → ACTIVATE(N)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
TAKE(s(M), cons(N, IL)) → U211(tt, activate(IL), M, N)
U211(tt, IL, M, N) → U221(tt, activate(IL), activate(M), activate(N))
U221(tt, IL, M, N) → ACTIVATE(IL)
U221(tt, IL, M, N) → ACTIVATE(M)
U221(tt, IL, M, N) → ACTIVATE(N)
U211(tt, IL, M, N) → ACTIVATE(IL)
U211(tt, IL, M, N) → ACTIVATE(M)
U211(tt, IL, M, N) → ACTIVATE(N)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
U231(tt, IL, M, N) → ACTIVATE(M)
U231(tt, IL, M, N) → ACTIVATE(IL)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
take(X1, X2) → n__take(X1, X2)
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(10) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

TAKE(s(M), cons(N, IL)) → U211(tt, activate(IL), M, N)
TAKE(s(M), cons(N, IL)) → ACTIVATE(IL)
The following rules are removed from R:

take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACTIVATE(x1)) = x1   
POL(TAKE(x1, x2)) = x1 + x2   
POL(U21(x1, x2, x3, x4)) = x1 + 2·x2 + x3 + 2·x4   
POL(U211(x1, x2, x3, x4)) = 2·x1 + x2 + x3 + x4   
POL(U22(x1, x2, x3, x4)) = x1 + 2·x2 + x3 + x4   
POL(U221(x1, x2, x3, x4)) = 2·x1 + x2 + x3 + x4   
POL(U23(x1, x2, x3, x4)) = 2·x1 + 2·x2 + x3 + x4   
POL(U231(x1, x2, x3, x4)) = 2·x1 + x2 + x3 + x4   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(n__take(x1, x2)) = x1 + 2·x2   
POL(n__zeros) = 0   
POL(s(x1)) = x1   
POL(take(x1, x2)) = x1 + 2·x2   
POL(tt) = 0   
POL(zeros) = 0   

(11) Obligation:

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

U221(tt, IL, M, N) → U231(tt, activate(IL), activate(M), activate(N))
U231(tt, IL, M, N) → ACTIVATE(N)
ACTIVATE(n__take(X1, X2)) → TAKE(X1, X2)
U211(tt, IL, M, N) → U221(tt, activate(IL), activate(M), activate(N))
U221(tt, IL, M, N) → ACTIVATE(IL)
U221(tt, IL, M, N) → ACTIVATE(M)
U221(tt, IL, M, N) → ACTIVATE(N)
U211(tt, IL, M, N) → ACTIVATE(IL)
U211(tt, IL, M, N) → ACTIVATE(M)
U211(tt, IL, M, N) → ACTIVATE(N)
U231(tt, IL, M, N) → ACTIVATE(M)
U231(tt, IL, M, N) → ACTIVATE(IL)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
take(X1, X2) → n__take(X1, X2)
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
zeros → cons(0, n__zeros)
zeros → n__zeros

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 0 SCCs with 12 less nodes.

(13) TRUE

(14) Obligation:

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

U121(tt, L) → LENGTH(activate(L))
LENGTH(cons(N, L)) → U111(tt, activate(L))
U111(tt, L) → U121(tt, activate(L))

The TRS R consists of the following rules:

zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
length(cons(N, L)) → U11(tt, activate(L))
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
zeros → n__zeros
take(X1, X2) → n__take(X1, X2)
activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → 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:

U121(tt, L) → LENGTH(activate(L))
LENGTH(cons(N, L)) → U111(tt, activate(L))
U111(tt, L) → U121(tt, activate(L))

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
take(X1, X2) → n__take(X1, X2)
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(17) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

take(s(M), cons(N, IL)) → U21(tt, activate(IL), M, N)
U21(tt, IL, M, N) → U22(tt, activate(IL), activate(M), activate(N))
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(0) = 0   
POL(LENGTH(x1)) = 2·x1   
POL(U111(x1, x2)) = 2·x1 + 2·x2   
POL(U121(x1, x2)) = x1 + 2·x2   
POL(U21(x1, x2, x3, x4)) = 2 + 2·x1 + 2·x2 + x3 + x4   
POL(U22(x1, x2, x3, x4)) = 2·x1 + 2·x2 + x3 + x4   
POL(U23(x1, x2, x3, x4)) = 2·x1 + 2·x2 + x3 + x4   
POL(activate(x1)) = x1   
POL(cons(x1, x2)) = x1 + x2   
POL(n__take(x1, x2)) = x1 + 2·x2   
POL(n__zeros) = 0   
POL(s(x1)) = 2 + 2·x1   
POL(take(x1, x2)) = x1 + 2·x2   
POL(tt) = 0   
POL(zeros) = 0   

(18) Obligation:

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

U121(tt, L) → LENGTH(activate(L))
LENGTH(cons(N, L)) → U111(tt, activate(L))
U111(tt, L) → U121(tt, activate(L))

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
take(X1, X2) → n__take(X1, X2)
U22(tt, IL, M, N) → U23(tt, activate(IL), activate(M), activate(N))
U23(tt, IL, M, N) → cons(activate(N), n__take(activate(M), activate(IL)))
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(19) 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.

(20) Obligation:

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

U121(tt, L) → LENGTH(activate(L))
LENGTH(cons(N, L)) → U111(tt, activate(L))
U111(tt, L) → U121(tt, activate(L))

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(21) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule U121(tt, L) → LENGTH(activate(L)) at position [0] we obtained the following new rules [LPAR04]:

U121(tt, n__zeros) → LENGTH(zeros)
U121(tt, n__take(x0, x1)) → LENGTH(take(x0, x1))
U121(tt, x0) → LENGTH(x0)

(22) Obligation:

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

LENGTH(cons(N, L)) → U111(tt, activate(L))
U111(tt, L) → U121(tt, activate(L))
U121(tt, n__zeros) → LENGTH(zeros)
U121(tt, n__take(x0, x1)) → LENGTH(take(x0, x1))
U121(tt, x0) → LENGTH(x0)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(23) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule U121(tt, n__zeros) → LENGTH(zeros) at position [0] we obtained the following new rules [LPAR04]:

U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, n__zeros) → LENGTH(n__zeros)

(24) Obligation:

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

LENGTH(cons(N, L)) → U111(tt, activate(L))
U111(tt, L) → U121(tt, activate(L))
U121(tt, n__take(x0, x1)) → LENGTH(take(x0, x1))
U121(tt, x0) → LENGTH(x0)
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, n__zeros) → LENGTH(n__zeros)

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(25) DependencyGraphProof (EQUIVALENT transformation)

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

(26) Obligation:

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

U111(tt, L) → U121(tt, activate(L))
U121(tt, n__take(x0, x1)) → LENGTH(take(x0, x1))
LENGTH(cons(N, L)) → U111(tt, activate(L))
U121(tt, x0) → LENGTH(x0)
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(27) Narrowing (EQUIVALENT transformation)

By narrowing [LPAR04] the rule U121(tt, n__take(x0, x1)) → LENGTH(take(x0, x1)) at position [0] we obtained the following new rules [LPAR04]:

U121(tt, n__take(x0, x1)) → LENGTH(n__take(x0, x1))

(28) Obligation:

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

U111(tt, L) → U121(tt, activate(L))
LENGTH(cons(N, L)) → U111(tt, activate(L))
U121(tt, x0) → LENGTH(x0)
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, n__take(x0, x1)) → LENGTH(n__take(x0, x1))

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(29) DependencyGraphProof (EQUIVALENT transformation)

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

(30) Obligation:

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

U121(tt, x0) → LENGTH(x0)
LENGTH(cons(N, L)) → U111(tt, activate(L))
U111(tt, L) → U121(tt, activate(L))
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))

The TRS R consists of the following rules:

activate(n__zeros) → zeros
activate(n__take(X1, X2)) → take(X1, X2)
activate(X) → X
take(X1, X2) → n__take(X1, X2)
zeros → cons(0, n__zeros)
zeros → n__zeros

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

(31) NonTerminationProof (EQUIVALENT 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 = U111(tt, activate(n__zeros)) evaluates to t =U111(tt, activate(n__zeros))

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



Rewriting sequence

U111(tt, activate(n__zeros)) → U111(tt, n__zeros)
with rule activate(X) → X at position [1] and matcher [X / n__zeros]

U111(tt, n__zeros) → U121(tt, activate(n__zeros))
with rule U111(tt, L) → U121(tt, activate(L)) at position [] and matcher [L / n__zeros]

U121(tt, activate(n__zeros)) → U121(tt, n__zeros)
with rule activate(X) → X at position [1] and matcher [X / n__zeros]

U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
with rule U121(tt, n__zeros) → LENGTH(cons(0, n__zeros)) at position [] and matcher [ ]

LENGTH(cons(0, n__zeros)) → U111(tt, activate(n__zeros))
with rule LENGTH(cons(N, L)) → U111(tt, activate(L))

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.



(32) FALSE