details

property	value
status	complete
benchmark	Ex4_4_Luc96b_C.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n013.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.604452848434 seconds
cpu usage	0.106235555
max memory	2977792.0

stage attributes

key	value
output-size	7616
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. active : [o] --> o f : [o * o] --> o g : [o] --> o mark : [o] --> o ok : [o] --> o proper : [o] --> o top : [o] --> o active(f(g(X), Y)) => mark(f(X, f(g(X), Y))) active(f(X, Y)) => f(active(X), Y) active(g(X)) => g(active(X)) f(mark(X), Y) => mark(f(X, Y)) g(mark(X)) => mark(g(X)) proper(f(X, Y)) => f(proper(X), proper(Y)) proper(g(X)) => g(proper(X)) f(ok(X), ok(Y)) => ok(f(X, Y)) g(ok(X)) => ok(g(X)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(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]): active(f(g(X), Y)) >? mark(f(X, f(g(X), Y))) active(f(X, Y)) >? f(active(X), Y) active(g(X)) >? g(active(X)) f(mark(X), Y) >? mark(f(X, Y)) g(mark(X)) >? mark(g(X)) proper(f(X, Y)) >? f(proper(X), proper(Y)) proper(g(X)) >? g(proper(X)) f(ok(X), ok(Y)) >? ok(f(X, Y)) g(ok(X)) >? ok(g(X)) top(mark(X)) >? top(proper(X)) top(ok(X)) >? top(active(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: active = \y0.1 + 2y0 f = \y0y1.y0 + y1 g = \y0.y0 mark = \y0.y0 ok = \y0.2 + 2y0 proper = \y0.y0 top = \y0.2y0 Using this interpretation, the requirements translate to: [[active(f(g(_x0), _x1))]] = 1 + 2x0 + 2x1 > x1 + 2x0 = [[mark(f(_x0, f(g(_x0), _x1)))]] [[active(f(_x0, _x1))]] = 1 + 2x0 + 2x1 >= 1 + x1 + 2x0 = [[f(active(_x0), _x1)]] [[active(g(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[g(active(_x0))]] [[f(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[mark(f(_x0, _x1))]] [[g(mark(_x0))]] = x0 >= x0 = [[mark(g(_x0))]] [[proper(f(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[f(proper(_x0), proper(_x1))]] [[proper(g(_x0))]] = x0 >= x0 = [[g(proper(_x0))]] [[f(ok(_x0), ok(_x1))]] = 4 + 2x0 + 2x1 > 2 + 2x0 + 2x1 = [[ok(f(_x0, _x1))]] [[g(ok(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[ok(g(_x0))]] [[top(mark(_x0))]] = 2x0 >= 2x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = 4 + 4x0 > 2 + 4x0 = [[top(active(_x0))]] We can thus remove the following rules: active(f(g(X), Y)) => mark(f(X, f(g(X), Y))) f(ok(X), ok(Y)) => ok(f(X, Y)) top(ok(X)) => top(active(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]): active(f(X, Y)) >? f(active(X), Y) active(g(X)) >? g(active(X)) f(mark(X), Y) >? mark(f(X, Y)) g(mark(X)) >? mark(g(X)) proper(f(X, Y)) >? f(proper(X), proper(Y)) proper(g(X)) >? g(proper(X)) g(ok(X)) >? ok(g(X)) top(mark(X)) >? top(proper(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: active = \y0.3y0 f = \y0y1.y0 + y1 g = \y0.2y0 mark = \y0.y0 popout

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