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TRS Stand 20472 pair #381715975
details
property
value
status
complete
benchmark
PEANO_nosorts_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n102.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.154795885086 seconds
cpu usage
0.148822712
max memory
3444736.0
stage attributes
key
value
output-size
11600
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!6220and : [o * o] --> o a!6220!6220plus : [o * o] --> o and : [o * o] --> o mark : [o] --> o plus : [o * o] --> o s : [o] --> o tt : [] --> o a!6220!6220and(tt, X) => mark(X) a!6220!6220plus(X, 0) => mark(X) a!6220!6220plus(X, s(Y)) => s(a!6220!6220plus(mark(X), mark(Y))) mark(and(X, Y)) => a!6220!6220and(mark(X), Y) mark(plus(X, Y)) => a!6220!6220plus(mark(X), mark(Y)) mark(tt) => tt mark(0) => 0 mark(s(X)) => s(mark(X)) a!6220!6220and(X, Y) => and(X, Y) a!6220!6220plus(X, Y) => plus(X, Y) 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!6220and(tt, X) >? mark(X) a!6220!6220plus(X, 0) >? mark(X) a!6220!6220plus(X, s(Y)) >? s(a!6220!6220plus(mark(X), mark(Y))) mark(and(X, Y)) >? a!6220!6220and(mark(X), Y) mark(plus(X, Y)) >? a!6220!6220plus(mark(X), mark(Y)) mark(tt) >? tt mark(0) >? 0 mark(s(X)) >? s(mark(X)) a!6220!6220and(X, Y) >? and(X, Y) a!6220!6220plus(X, Y) >? plus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 a!6220!6220and = \y0y1.y1 + 2y0 a!6220!6220plus = \y0y1.y0 + y1 and = \y0y1.y1 + 2y0 mark = \y0.y0 plus = \y0y1.y0 + y1 s = \y0.y0 tt = 0 Using this interpretation, the requirements translate to: [[a!6220!6220and(tt, _x0)]] = x0 >= x0 = [[mark(_x0)]] [[a!6220!6220plus(_x0, 0)]] = 1 + x0 > x0 = [[mark(_x0)]] [[a!6220!6220plus(_x0, s(_x1))]] = x0 + x1 >= x0 + x1 = [[s(a!6220!6220plus(mark(_x0), mark(_x1)))]] [[mark(and(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[a!6220!6220and(mark(_x0), _x1)]] [[mark(plus(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[a!6220!6220plus(mark(_x0), mark(_x1))]] [[mark(tt)]] = 0 >= 0 = [[tt]] [[mark(0)]] = 1 >= 1 = [[0]] [[mark(s(_x0))]] = x0 >= x0 = [[s(mark(_x0))]] [[a!6220!6220and(_x0, _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[and(_x0, _x1)]] [[a!6220!6220plus(_x0, _x1)]] = x0 + x1 >= x0 + x1 = [[plus(_x0, _x1)]] We can thus remove the following rules: a!6220!6220plus(X, 0) => mark(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!6220and(tt, X) >? mark(X) a!6220!6220plus(X, s(Y)) >? s(a!6220!6220plus(mark(X), mark(Y))) mark(and(X, Y)) >? a!6220!6220and(mark(X), Y) mark(plus(X, Y)) >? a!6220!6220plus(mark(X), mark(Y)) mark(tt) >? tt mark(0) >? 0 mark(s(X)) >? s(mark(X)) a!6220!6220and(X, Y) >? and(X, Y) a!6220!6220plus(X, Y) >? plus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 a!6220!6220and = \y0y1.y0 + y1 a!6220!6220plus = \y0y1.y0 + y1 and = \y0y1.y0 + y1 mark = \y0.y0 plus = \y0y1.y0 + y1
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