YES
proof of SK90_2.26.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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, 65 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:

   f(0) -> 0
   f(s(0)) -> s(0)
   f(s(s(x))) -> p(h(g(x)))
   g(0) -> pair(s(0), s(0))
   g(s(x)) -> h(g(x))
   h(x) -> pair(+(p(x), q(x)), p(x))
   p(pair(x, y)) -> x
   q(pair(x, y)) -> y
   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   f(s(s(x))) -> +(p(g(x)), q(g(x)))
   g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
f/1(YES)
0/0)
s/1(YES)
p/1)YES(
h/1(YES)
g/1(YES)
pair/2(YES,YES)
+/2(YES,YES)
q/1)YES(

Quasi precedence:
f_1 > g_1 > 0 > s_1 > pair_2
f_1 > g_1 > h_1 > +_2 > s_1 > pair_2


Status:
f_1: [1]
0: multiset status
s_1: multiset status
h_1: [1]
g_1: [1]
pair_2: [1,2]
+_2: [2,1]

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

   f(0) -> 0
   f(s(0)) -> s(0)
   f(s(s(x))) -> p(h(g(x)))
   g(0) -> pair(s(0), s(0))
   g(s(x)) -> h(g(x))
   h(x) -> pair(+(p(x), q(x)), p(x))
   p(pair(x, y)) -> x
   q(pair(x, y)) -> y
   +(x, 0) -> x
   +(x, s(y)) -> s(+(x, y))
   f(s(s(x))) -> +(p(g(x)), q(g(x)))
   g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(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