details

property	value
status	complete
benchmark	LPAR_intlist.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n108.star.cs.uiowa.edu
space	AProVE_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.111233949661 seconds
cpu usage	0.101079622
max memory	4038656.0

stage attributes

key	value
output-size	6549
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o cons : [o * o] --> o int : [o * o] --> o intlist : [o] --> o nil : [] --> o s : [o] --> o intlist(nil) => nil int(s(X), 0) => nil int(X, X) => cons(X, nil) intlist(cons(X, Y)) => cons(s(X), intlist(Y)) int(s(X), s(Y)) => intlist(int(X, Y)) int(0, s(X)) => cons(0, int(s(0), s(X))) intlist(cons(X, nil)) => cons(s(X), nil) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): intlist(nil) >? nil int(s(X), 0) >? nil int(X, X) >? cons(X, nil) intlist(cons(X, Y)) >? cons(s(X), intlist(Y)) int(s(X), s(Y)) >? intlist(int(X, Y)) int(0, s(X)) >? cons(0, int(s(0), s(X))) intlist(cons(X, nil)) >? cons(s(X), nil) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.y0 + y1 int = \y0y1.1 + y0 + y1 intlist = \y0.y0 nil = 0 s = \y0.y0 Using this interpretation, the requirements translate to: [[intlist(nil)]] = 0 >= 0 = [[nil]] [[int(s(_x0), 0)]] = 1 + x0 > 0 = [[nil]] [[int(_x0, _x0)]] = 1 + 2x0 > x0 = [[cons(_x0, nil)]] [[intlist(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[cons(s(_x0), intlist(_x1))]] [[int(s(_x0), s(_x1))]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[intlist(int(_x0, _x1))]] [[int(0, s(_x0))]] = 1 + x0 >= 1 + x0 = [[cons(0, int(s(0), s(_x0)))]] [[intlist(cons(_x0, nil))]] = x0 >= x0 = [[cons(s(_x0), nil)]] We can thus remove the following rules: int(s(X), 0) => nil int(X, X) => cons(X, nil) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): intlist(nil) >? nil intlist(cons(X, Y)) >? cons(s(X), intlist(Y)) int(s(X), s(Y)) >? intlist(int(X, Y)) int(0, s(X)) >? cons(0, int(s(0), s(X))) intlist(cons(X, nil)) >? cons(s(X), nil) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.y0 + y1 int = \y0y1.y0 + y1 intlist = \y0.2y0 nil = 1 s = \y0.2y0 Using this interpretation, the requirements translate to: [[intlist(nil)]] = 2 > 1 = [[nil]] [[intlist(cons(_x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[cons(s(_x0), intlist(_x1))]] [[int(s(_x0), s(_x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[intlist(int(_x0, _x1))]] [[int(0, s(_x0))]] = 2x0 >= 2x0 = [[cons(0, int(s(0), s(_x0)))]] [[intlist(cons(_x0, nil))]] = 2 + 2x0 > 1 + 2x0 = [[cons(s(_x0), nil)]] We can thus remove the following rules: intlist(nil) => nil intlist(cons(X, nil)) => cons(s(X), nil) We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). popout

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