YES Termination w.r.t. Q proof of Transformed_CSR_04_ExSec4_2_DLMMU04_Z.ari

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

0 QTRS
↳1 QTRSRRRProof (⇔, 74 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 RisEmptyProof (⇔, 0 ms)
↳6 YES

(0) Obligation:

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

natsFrom(N) → cons(N, n__natsFrom(s(N)))
fst(pair(XS, YS)) → XS
snd(pair(XS, YS)) → YS
splitAt(0, XS) → pair(nil, XS)
splitAt(s(N), cons(X, XS)) → u(splitAt(N, activate(XS)), N, X, activate(XS))
u(pair(YS, ZS), N, X, XS) → pair(cons(activate(X), YS), ZS)
head(cons(N, XS)) → N
tail(cons(N, XS)) → activate(XS)
sel(N, XS) → head(afterNth(N, XS))
take(N, XS) → fst(splitAt(N, XS))
afterNth(N, XS) → snd(splitAt(N, XS))
natsFrom(X) → n__natsFrom(X)
activate(n__natsFrom(X)) → natsFrom(X)
activate(X) → X

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
natsFrom(x1)  =  natsFrom(x1)
cons(x1, x2)  =  cons(x1, x2)
n__natsFrom(x1)  =  x1
s(x1)  =  s(x1)
fst(x1)  =  x1
pair(x1, x2)  =  pair(x1, x2)
snd(x1)  =  snd(x1)
splitAt(x1, x2)  =  splitAt(x1, x2)
0  =  0
nil  =  nil
u(x1, x2, x3, x4)  =  u(x1, x2, x3, x4)
activate(x1)  =  activate(x1)
head(x1)  =  x1
tail(x1)  =  tail(x1)
sel(x1, x2)  =  sel(x1, x2)
afterNth(x1, x2)  =  afterNth(x1, x2)
take(x1, x2)  =  take(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:
tail1 > [natsFrom1, activate1] > cons2 > [snd1, 0, nil]
tail1 > [natsFrom1, activate1] > s1 > [snd1, 0, nil]
[sel2, afterNth2] > splitAt2 > u4 > [natsFrom1, activate1] > cons2 > [snd1, 0, nil]
[sel2, afterNth2] > splitAt2 > u4 > [natsFrom1, activate1] > s1 > [snd1, 0, nil]
[sel2, afterNth2] > splitAt2 > u4 > pair2 > [snd1, 0, nil]
take2 > splitAt2 > u4 > [natsFrom1, activate1] > cons2 > [snd1, 0, nil]
take2 > splitAt2 > u4 > [natsFrom1, activate1] > s1 > [snd1, 0, nil]
take2 > splitAt2 > u4 > pair2 > [snd1, 0, nil]

Status:
natsFrom1: [1]
cons2: [1,2]
s1: [1]
pair2: multiset
snd1: [1]
splitAt2: [1,2]
0: multiset
nil: multiset
u4: multiset
activate1: [1]
tail1: [1]
sel2: multiset
afterNth2: multiset
take2: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

natsFrom(N) → cons(N, n__natsFrom(s(N)))
fst(pair(XS, YS)) → XS
snd(pair(XS, YS)) → YS
splitAt(0, XS) → pair(nil, XS)
splitAt(s(N), cons(X, XS)) → u(splitAt(N, activate(XS)), N, X, activate(XS))
u(pair(YS, ZS), N, X, XS) → pair(cons(activate(X), YS), ZS)
head(cons(N, XS)) → N
tail(cons(N, XS)) → activate(XS)
take(N, XS) → fst(splitAt(N, XS))
afterNth(N, XS) → snd(splitAt(N, XS))
natsFrom(X) → n__natsFrom(X)
activate(X) → X


(2) Obligation:

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

sel(N, XS) → head(afterNth(N, XS))
activate(n__natsFrom(X)) → natsFrom(X)

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Knuth-Bendix order [KBO] with precedence:
activate1 > sel2 > natsFrom1 > nnatsFrom1 > afterNth2 > head1

and weight map:

head_1=1
activate_1=1
n__natsFrom_1=1
natsFrom_1=2
sel_2=1
afterNth_2=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

sel(N, XS) → head(afterNth(N, XS))
activate(n__natsFrom(X)) → natsFrom(X)


(4) Obligation:

Q restricted rewrite system:
R is empty.
Q is empty.

(5) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.

(6) YES