YES
proof of Secret_07_TRS_secret3.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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, 57 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 0 ms]
(4) QTRS
(5) RisEmptyProof [EQUIVALENT, 0 ms]
(6) YES


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

   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, nil)) -> cons(x, nil)
   sum(cons(x, cons(y, l))) -> sum(cons(a(x, y, h), l))
   a(h, h, x) -> s(x)
   a(x, s(y), h) -> a(x, y, s(h))
   a(x, s(y), s(z)) -> a(x, y, a(x, s(y), z))
   a(s(x), h, z) -> a(x, z, z)

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
app/2(YES,YES)
nil/0)
cons/2(YES,YES)
sum/1)YES(
a/3(YES,YES,YES)
h/0)
s/1(YES)

Quasi precedence:
app_2 > [nil, cons_2, h] > a_3 > s_1


Status:
app_2: multiset status
nil: multiset status
cons_2: [2,1]
a_3: [1,2,3]
h: multiset status
s_1: multiset status

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

   app(nil, k) -> k
   app(l, nil) -> l
   app(cons(x, l), k) -> cons(x, app(l, k))
   sum(cons(x, cons(y, l))) -> sum(cons(a(x, y, h), l))
   a(h, h, x) -> s(x)
   a(x, s(y), h) -> a(x, y, s(h))
   a(x, s(y), s(z)) -> a(x, y, a(x, s(y), z))
   a(s(x), h, z) -> a(x, z, z)




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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   sum(cons(x, nil)) -> cons(x, nil)

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:sum_1 > nil > cons_2

and weight map:

   nil=1
   sum_1=0
   cons_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:

   sum(cons(x, nil)) -> cons(x, nil)




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(4)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(5) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(6)
YES