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TRS Stand 20472 pair #381711858
details
property
value
status
complete
benchmark
ReverseLastInit.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n080.star.cs.uiowa.edu
space
Applicative_05
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.120558977127 seconds
cpu usage
0.117076509
max memory
3420160.0
stage attributes
key
value
output-size
5974
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. app : [o * o] --> o compose : [] --> o cons : [] --> o hd : [] --> o init : [] --> o last : [] --> o nil : [] --> o reverse : [] --> o reverse2 : [] --> o tl : [] --> o app(app(app(compose, X), Y), Z) => app(Y, app(X, Z)) app(reverse, X) => app(app(reverse2, X), nil) app(app(reverse2, nil), X) => X app(app(reverse2, app(app(cons, X), Y)), Z) => app(app(reverse2, Y), app(app(cons, X), Z)) app(hd, app(app(cons, X), Y)) => X app(tl, app(app(cons, X), Y)) => Y last => app(app(compose, hd), reverse) init => app(app(compose, reverse), app(app(compose, tl), reverse)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): app(app(app(compose, X), Y), Z) >? app(Y, app(X, Z)) app(reverse, X) >? app(app(reverse2, X), nil) app(app(reverse2, nil), X) >? X app(app(reverse2, app(app(cons, X), Y)), Z) >? app(app(reverse2, Y), app(app(cons, X), Z)) app(hd, app(app(cons, X), Y)) >? X app(tl, app(app(cons, X), Y)) >? Y last >? app(app(compose, hd), reverse) init >? app(app(compose, reverse), app(app(compose, tl), reverse)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: app = \y0y1.y0 + y1 compose = 0 cons = 0 hd = 1 init = 3 last = 3 nil = 0 reverse = 0 reverse2 = 0 tl = 0 Using this interpretation, the requirements translate to: [[app(app(app(compose, _x0), _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[app(_x1, app(_x0, _x2))]] [[app(reverse, _x0)]] = x0 >= x0 = [[app(app(reverse2, _x0), nil)]] [[app(app(reverse2, nil), _x0)]] = x0 >= x0 = [[_x0]] [[app(app(reverse2, app(app(cons, _x0), _x1)), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[app(app(reverse2, _x1), app(app(cons, _x0), _x2))]] [[app(hd, app(app(cons, _x0), _x1))]] = 1 + x0 + x1 > x0 = [[_x0]] [[app(tl, app(app(cons, _x0), _x1))]] = x0 + x1 >= x1 = [[_x1]] [[last]] = 3 > 1 = [[app(app(compose, hd), reverse)]] [[init]] = 3 > 0 = [[app(app(compose, reverse), app(app(compose, tl), reverse))]] We can thus remove the following rules: app(hd, app(app(cons, X), Y)) => X last => app(app(compose, hd), reverse) init => app(app(compose, reverse), app(app(compose, tl), reverse)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): app(app(app(compose, X), Y), Z) >? app(Y, app(X, Z)) app(reverse, X) >? app(app(reverse2, X), nil) app(app(reverse2, nil), X) >? X app(app(reverse2, app(app(cons, X), Y)), Z) >? app(app(reverse2, Y), app(app(cons, X), Z)) app(tl, app(app(cons, X), Y)) >? Y We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: app = \y0y1.y0 + y1 compose = 3 cons = 0 nil = 0 reverse = 3 reverse2 = 0 tl = 3 Using this interpretation, the requirements translate to:
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