TRS Standard pair #487068973

details

property	value
status	complete
benchmark	identity.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n179.star.cs.uiowa.edu
space	AProVE_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.26532 seconds
cpu usage	5.89847
user time	5.64348
system time	0.254986
max virtual memory	1.8477524E7
max residence set size	379584.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 46 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 17 ms]
(4) QTRS
(5) AAECC Innermost [EQUIVALENT, 0 ms]
(6) QTRS
(7) DependencyPairsProof [EQUIVALENT, 2 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) AND
    (11) QDP
        (12) UsableRulesProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QReductionProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (17) YES
    (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QReductionProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES
    (25) QDP
        (26) UsableRulesProof [EQUIVALENT, 0 ms]
        (27) QDP
        (28) QReductionProof [EQUIVALENT, 0 ms]
        (29) QDP
        (30) QDPOrderProof [EQUIVALENT, 45 ms]
        (31) QDP
        (32) PisEmptyProof [EQUIVALENT, 0 ms]
        (33) YES


----------------------------------------

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

   g(x, 0) -> 0
   g(d, s(x)) -> s(s(g(d, x)))
   g(h, s(0)) -> 0
   g(h, s(s(x))) -> s(g(h, x))
   double(x) -> g(d, x)
   half(x) -> g(h, x)
   f(s(x), y) -> f(half(s(x)), double(y))
   f(s(0), y) -> y
   id(x) -> f(x, s(0))

Q is empty.

----------------------------------------

(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(d) = 0
   POL(double(x_1)) = x_1
   POL(f(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(g(x_1, x_2)) = x_1 + x_2
   POL(h) = 0
   POL(half(x_1)) = x_1
   POL(id(x_1)) = 2 + 2*x_1
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   id(x) -> f(x, s(0))




----------------------------------------

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

   g(x, 0) -> 0
   g(d, s(x)) -> s(s(g(d, x)))
   g(h, s(0)) -> 0
   g(h, s(s(x))) -> s(g(h, x))
   double(x) -> g(d, x)
   half(x) -> g(h, x)
   f(s(x), y) -> f(half(s(x)), double(y))
   f(s(0), y) -> y

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