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TRS Stand 20472 pair #381711881
details
property
value
status
complete
benchmark
ExConc_Zan97_C.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n111.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0826258659363 seconds
cpu usage
0.079656763
max memory
2539520.0
stage attributes
key
value
output-size
6213
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. active : [o] --> o f : [o] --> o g : [o] --> o h : [o] --> o mark : [o] --> o ok : [o] --> o proper : [o] --> o top : [o] --> o active(f(X)) => mark(g(h(f(X)))) active(f(X)) => f(active(X)) active(h(X)) => h(active(X)) f(mark(X)) => mark(f(X)) h(mark(X)) => mark(h(X)) proper(f(X)) => f(proper(X)) proper(g(X)) => g(proper(X)) proper(h(X)) => h(proper(X)) f(ok(X)) => ok(f(X)) g(ok(X)) => ok(g(X)) h(ok(X)) => ok(h(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(X)) >? mark(g(h(f(X)))) active(f(X)) >? f(active(X)) active(h(X)) >? h(active(X)) f(mark(X)) >? mark(f(X)) h(mark(X)) >? mark(h(X)) proper(f(X)) >? f(proper(X)) proper(g(X)) >? g(proper(X)) proper(h(X)) >? h(proper(X)) f(ok(X)) >? ok(f(X)) g(ok(X)) >? ok(g(X)) h(ok(X)) >? ok(h(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 = \y0.2 + 3y0 f = \y0.2 + 3y0 g = \y0.y0 h = \y0.y0 mark = \y0.1 + 2y0 ok = \y0.2 + 3y0 proper = \y0.y0 top = \y0.y0 Using this interpretation, the requirements translate to: [[active(f(_x0))]] = 8 + 9x0 > 5 + 6x0 = [[mark(g(h(f(_x0))))]] [[active(f(_x0))]] = 8 + 9x0 >= 8 + 9x0 = [[f(active(_x0))]] [[active(h(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[h(active(_x0))]] [[f(mark(_x0))]] = 5 + 6x0 >= 5 + 6x0 = [[mark(f(_x0))]] [[h(mark(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark(h(_x0))]] [[proper(f(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[f(proper(_x0))]] [[proper(g(_x0))]] = x0 >= x0 = [[g(proper(_x0))]] [[proper(h(_x0))]] = x0 >= x0 = [[h(proper(_x0))]] [[f(ok(_x0))]] = 8 + 9x0 >= 8 + 9x0 = [[ok(f(_x0))]] [[g(ok(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[ok(g(_x0))]] [[h(ok(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[ok(h(_x0))]] [[top(mark(_x0))]] = 1 + 2x0 > x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = 2 + 3x0 >= 2 + 3x0 = [[top(active(_x0))]] We can thus remove the following rules: active(f(X)) => mark(g(h(f(X)))) top(mark(X)) => top(proper(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)) >? f(active(X)) active(h(X)) >? h(active(X)) f(mark(X)) >? mark(f(X)) h(mark(X)) >? mark(h(X)) proper(f(X)) >? f(proper(X)) proper(g(X)) >? g(proper(X)) proper(h(X)) >? h(proper(X)) f(ok(X)) >? ok(f(X)) g(ok(X)) >? ok(g(X)) h(ok(X)) >? ok(h(X))
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