YES
proof of CiME_04_ack_prolog.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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 jera 20211004 unpublished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, 69 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:

   ack_in(0, n) -> ack_out(s(n))
   ack_in(s(m), 0) -> u11(ack_in(m, s(0)))
   u11(ack_out(n)) -> ack_out(n)
   ack_in(s(m), s(n)) -> u21(ack_in(s(m), n), m)
   u21(ack_out(n), m) -> u22(ack_in(m, n))
   u22(ack_out(n)) -> ack_out(n)

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
ack_in/2(YES,YES)
0/0)
ack_out/1(YES)
s/1(YES)
u11/1)YES(
u21/2(YES,YES)
u22/1)YES(

Quasi precedence:
[ack_in_2, u21_2] > 0 > ack_out_1
[ack_in_2, u21_2] > 0 > s_1


Status:
ack_in_2: [1,2]
0: multiset status
ack_out_1: multiset status
s_1: multiset status
u21_2: [2,1]

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

   ack_in(0, n) -> ack_out(s(n))
   ack_in(s(m), 0) -> u11(ack_in(m, s(0)))
   ack_in(s(m), s(n)) -> u21(ack_in(s(m), n), m)
   u21(ack_out(n), m) -> u22(ack_in(m, n))




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

   u11(ack_out(n)) -> ack_out(n)
   u22(ack_out(n)) -> ack_out(n)

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:u22_1 > u11_1 > ack_out_1

and weight map:

   u11_1=1
   ack_out_1=1
   u22_1=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:

   u11(ack_out(n)) -> ack_out(n)
   u22(ack_out(n)) -> ack_out(n)




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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