YES
proof of Transformed_CSR_04_ExSec11_1_Luc02a_L.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 70 ms]
(2) QTRS
(3) RisEmptyProof [EQUIVALENT, 0 ms]
(4) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   terms(N) -> cons(recip(sqr(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y)) -> cons(Y)
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(X))) -> s(half(X))
   half(dbl(X)) -> X

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
terms/1(YES)
cons/1)YES(
recip/1)YES(
sqr/1(YES)
0/0)
s/1(YES)
add/2(YES,YES)
dbl/1(YES)
first/2(YES,YES)
nil/0)
half/1(YES)

Quasi precedence:
terms_1 > [sqr_1, dbl_1] > add_2 > s_1
0 > [first_2, nil] > s_1
half_1 > s_1


Status:
terms_1: multiset status
sqr_1: multiset status
0: multiset status
s_1: multiset status
add_2: multiset status
dbl_1: multiset status
first_2: [1,2]
nil: multiset status
half_1: multiset status

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

   terms(N) -> cons(recip(sqr(N)))
   sqr(0) -> 0
   sqr(s(X)) -> s(add(sqr(X), dbl(X)))
   dbl(0) -> 0
   dbl(s(X)) -> s(s(dbl(X)))
   add(0, X) -> X
   add(s(X), Y) -> s(add(X, Y))
   first(0, X) -> nil
   first(s(X), cons(Y)) -> cons(Y)
   half(0) -> 0
   half(s(0)) -> 0
   half(s(s(X))) -> s(half(X))
   half(dbl(X)) -> X




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(2)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(3) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(4)
YES