details

property	value
status	complete
benchmark	2.42.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	SK90

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0760171413422 seconds
cpu usage	0.072387534
max memory	2252800.0

stage attributes

key	value
output-size	6893
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. !plus!plus : [o * o] --> o flatten : [o] --> o nil : [] --> o rev : [o] --> o unit : [o] --> o flatten(nil) => nil flatten(unit(X)) => flatten(X) flatten(!plus!plus(X, Y)) => !plus!plus(flatten(X), flatten(Y)) flatten(!plus!plus(unit(X), Y)) => !plus!plus(flatten(X), flatten(Y)) flatten(flatten(X)) => flatten(X) rev(nil) => nil rev(unit(X)) => unit(X) rev(!plus!plus(X, Y)) => !plus!plus(rev(Y), rev(X)) rev(rev(X)) => X !plus!plus(X, nil) => X !plus!plus(nil, X) => X !plus!plus(!plus!plus(X, Y), Z) => !plus!plus(X, !plus!plus(Y, Z)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): flatten(nil) >? nil flatten(unit(X)) >? flatten(X) flatten(!plus!plus(X, Y)) >? !plus!plus(flatten(X), flatten(Y)) flatten(!plus!plus(unit(X), Y)) >? !plus!plus(flatten(X), flatten(Y)) flatten(flatten(X)) >? flatten(X) rev(nil) >? nil rev(unit(X)) >? unit(X) rev(!plus!plus(X, Y)) >? !plus!plus(rev(Y), rev(X)) rev(rev(X)) >? X !plus!plus(X, nil) >? X !plus!plus(nil, X) >? X !plus!plus(!plus!plus(X, Y), Z) >? !plus!plus(X, !plus!plus(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus!plus = \y0y1.y0 + y1 flatten = \y0.y0 nil = 0 rev = \y0.2y0 unit = \y0.2 + y0 Using this interpretation, the requirements translate to: [[flatten(nil)]] = 0 >= 0 = [[nil]] [[flatten(unit(_x0))]] = 2 + x0 > x0 = [[flatten(_x0)]] [[flatten(!plus!plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!plus!plus(flatten(_x0), flatten(_x1))]] [[flatten(!plus!plus(unit(_x0), _x1))]] = 2 + x0 + x1 > x0 + x1 = [[!plus!plus(flatten(_x0), flatten(_x1))]] [[flatten(flatten(_x0))]] = x0 >= x0 = [[flatten(_x0)]] [[rev(nil)]] = 0 >= 0 = [[nil]] [[rev(unit(_x0))]] = 4 + 2x0 > 2 + x0 = [[unit(_x0)]] [[rev(!plus!plus(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[!plus!plus(rev(_x1), rev(_x0))]] [[rev(rev(_x0))]] = 4x0 >= x0 = [[_x0]] [[!plus!plus(_x0, nil)]] = x0 >= x0 = [[_x0]] [[!plus!plus(nil, _x0)]] = x0 >= x0 = [[_x0]] [[!plus!plus(!plus!plus(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!plus!plus(_x0, !plus!plus(_x1, _x2))]] We can thus remove the following rules: flatten(unit(X)) => flatten(X) flatten(!plus!plus(unit(X), Y)) => !plus!plus(flatten(X), flatten(Y)) rev(unit(X)) => unit(X) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): flatten(nil) >? nil flatten(!plus!plus(X, Y)) >? !plus!plus(flatten(X), flatten(Y)) flatten(flatten(X)) >? flatten(X) rev(nil) >? nil rev(!plus!plus(X, Y)) >? !plus!plus(rev(Y), rev(X)) rev(rev(X)) >? X !plus!plus(X, nil) >? X !plus!plus(nil, X) >? X !plus!plus(!plus!plus(X, Y), Z) >? !plus!plus(X, !plus!plus(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus!plus = \y0y1.y0 + y1 flatten = \y0.2y0 nil = 2 rev = \y0.2y0 popout

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