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TRS Stand 20472 pair #381716515
details
property
value
status
complete
benchmark
Ex5_Zan97_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n024.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.267162084579 seconds
cpu usage
0.263675898
max memory
9003008.0
stage attributes
key
value
output-size
15005
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 c : [] --> o f : [o] --> o false : [] --> o if : [o * o * o] --> o mark : [o] --> o true : [] --> o active(f(X)) => mark(if(X, c, f(true))) active(if(true, X, Y)) => mark(X) active(if(false, X, Y)) => mark(Y) mark(f(X)) => active(f(mark(X))) mark(if(X, Y, Z)) => active(if(mark(X), mark(Y), Z)) mark(c) => active(c) mark(true) => active(true) mark(false) => active(false) f(mark(X)) => f(X) f(active(X)) => f(X) if(mark(X), Y, Z) => if(X, Y, Z) if(X, mark(Y), Z) => if(X, Y, Z) if(X, Y, mark(Z)) => if(X, Y, Z) if(active(X), Y, Z) => if(X, Y, Z) if(X, active(Y), Z) => if(X, Y, Z) if(X, Y, active(Z)) => if(X, 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]): active(f(X)) >? mark(if(X, c, f(true))) active(if(true, X, Y)) >? mark(X) active(if(false, X, Y)) >? mark(Y) mark(f(X)) >? active(f(mark(X))) mark(if(X, Y, Z)) >? active(if(mark(X), mark(Y), Z)) mark(c) >? active(c) mark(true) >? active(true) mark(false) >? active(false) f(mark(X)) >? f(X) f(active(X)) >? f(X) if(mark(X), Y, Z) >? if(X, Y, Z) if(X, mark(Y), Z) >? if(X, Y, Z) if(X, Y, mark(Z)) >? if(X, Y, Z) if(active(X), Y, Z) >? if(X, Y, Z) if(X, active(Y), Z) >? if(X, Y, Z) if(X, Y, active(Z)) >? if(X, Y, Z) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: active = \y0.y0 c = 0 f = \y0.3y0 false = 2 if = \y0y1y2.y0 + 2y1 + 3y2 mark = \y0.2y0 true = 0 Using this interpretation, the requirements translate to: [[active(f(_x0))]] = 3x0 >= 2x0 = [[mark(if(_x0, c, f(true)))]] [[active(if(true, _x0, _x1))]] = 2x0 + 3x1 >= 2x0 = [[mark(_x0)]] [[active(if(false, _x0, _x1))]] = 2 + 2x0 + 3x1 > 2x1 = [[mark(_x1)]] [[mark(f(_x0))]] = 6x0 >= 6x0 = [[active(f(mark(_x0)))]] [[mark(if(_x0, _x1, _x2))]] = 2x0 + 4x1 + 6x2 >= 2x0 + 3x2 + 4x1 = [[active(if(mark(_x0), mark(_x1), _x2))]] [[mark(c)]] = 0 >= 0 = [[active(c)]] [[mark(true)]] = 0 >= 0 = [[active(true)]] [[mark(false)]] = 4 > 2 = [[active(false)]] [[f(mark(_x0))]] = 6x0 >= 3x0 = [[f(_x0)]] [[f(active(_x0))]] = 3x0 >= 3x0 = [[f(_x0)]] [[if(mark(_x0), _x1, _x2)]] = 2x0 + 2x1 + 3x2 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] [[if(_x0, mark(_x1), _x2)]] = x0 + 3x2 + 4x1 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] [[if(_x0, _x1, mark(_x2))]] = x0 + 2x1 + 6x2 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] [[if(active(_x0), _x1, _x2)]] = x0 + 2x1 + 3x2 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] [[if(_x0, active(_x1), _x2)]] = x0 + 2x1 + 3x2 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] [[if(_x0, _x1, active(_x2))]] = x0 + 2x1 + 3x2 >= x0 + 2x1 + 3x2 = [[if(_x0, _x1, _x2)]] We can thus remove the following rules: active(if(false, X, Y)) => mark(Y) mark(false) => active(false) 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]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] active#(f(X)) =#> mark#(if(X, c, f(true)))
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