details

property	value
status	complete
benchmark	Ex1_GL02a_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n084.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.252833127975 seconds
cpu usage	0.24383874
max memory	8720384.0

stage attributes

key	value
output-size	15375
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 a!6220!6220eq : [o * o] --> o a!6220!6220inf : [o] --> o a!6220!6220length : [o] --> o a!6220!6220take : [o * o] --> o cons : [o * o] --> o eq : [o * o] --> o false : [] --> o inf : [o] --> o length : [o] --> o mark : [o] --> o nil : [] --> o s : [o] --> o take : [o * o] --> o true : [] --> o a!6220!6220eq(0, 0) => true a!6220!6220eq(s(X), s(Y)) => a!6220!6220eq(X, Y) a!6220!6220eq(X, Y) => false a!6220!6220inf(X) => cons(X, inf(s(X))) a!6220!6220take(0, X) => nil a!6220!6220take(s(X), cons(Y, Z)) => cons(Y, take(X, Z)) a!6220!6220length(nil) => 0 a!6220!6220length(cons(X, Y)) => s(length(Y)) mark(eq(X, Y)) => a!6220!6220eq(X, Y) mark(inf(X)) => a!6220!6220inf(mark(X)) mark(take(X, Y)) => a!6220!6220take(mark(X), mark(Y)) mark(length(X)) => a!6220!6220length(mark(X)) mark(0) => 0 mark(true) => true mark(s(X)) => s(X) mark(false) => false mark(cons(X, Y)) => cons(X, Y) mark(nil) => nil a!6220!6220eq(X, Y) => eq(X, Y) a!6220!6220inf(X) => inf(X) a!6220!6220take(X, Y) => take(X, Y) a!6220!6220length(X) => length(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]): a!6220!6220eq(0, 0) >? true a!6220!6220eq(s(X), s(Y)) >? a!6220!6220eq(X, Y) a!6220!6220eq(X, Y) >? false a!6220!6220inf(X) >? cons(X, inf(s(X))) a!6220!6220take(0, X) >? nil a!6220!6220take(s(X), cons(Y, Z)) >? cons(Y, take(X, Z)) a!6220!6220length(nil) >? 0 a!6220!6220length(cons(X, Y)) >? s(length(Y)) mark(eq(X, Y)) >? a!6220!6220eq(X, Y) mark(inf(X)) >? a!6220!6220inf(mark(X)) mark(take(X, Y)) >? a!6220!6220take(mark(X), mark(Y)) mark(length(X)) >? a!6220!6220length(mark(X)) mark(0) >? 0 mark(true) >? true mark(s(X)) >? s(X) mark(false) >? false mark(cons(X, Y)) >? cons(X, Y) mark(nil) >? nil a!6220!6220eq(X, Y) >? eq(X, Y) a!6220!6220inf(X) >? inf(X) a!6220!6220take(X, Y) >? take(X, Y) a!6220!6220length(X) >? length(X) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[a!6220!6220length(x_1)]] = x_1 [[false]] = _|_ [[length(x_1)]] = x_1 [[nil]] = _|_ [[true]] = _|_ We choose Lex = {} and Mul = {a!6220!6220eq, a!6220!6220inf, a!6220!6220take, cons, eq, inf, mark, s, take}, and the following precedence: mark > a!6220!6220inf > inf > a!6220!6220take = take > cons > a!6220!6220eq > eq > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: a!6220!6220eq(_|_, _|_) >= _|_ a!6220!6220eq(s(X), s(Y)) >= a!6220!6220eq(X, Y) a!6220!6220eq(X, Y) >= _|_ a!6220!6220inf(X) >= cons(X, inf(s(X))) a!6220!6220take(_|_, X) >= _|_ a!6220!6220take(s(X), cons(Y, Z)) > cons(Y, take(X, Z)) popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Stand 20472