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TRS Stand 20472 pair #381713379
details
property
value
status
complete
benchmark
ExProp7_Luc06_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n022.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.204905986786 seconds
cpu usage
0.153321713
max memory
3686400.0
stage attributes
key
value
output-size
9165
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. 0 : [] --> o a!6220!6220f : [o] --> o a!6220!6220p : [o] --> o cons : [o * o] --> o f : [o] --> o mark : [o] --> o p : [o] --> o s : [o] --> o a!6220!6220f(0) => cons(0, f(s(0))) a!6220!6220f(s(0)) => a!6220!6220f(a!6220!6220p(s(0))) a!6220!6220p(s(X)) => mark(X) mark(f(X)) => a!6220!6220f(mark(X)) mark(p(X)) => a!6220!6220p(mark(X)) mark(0) => 0 mark(cons(X, Y)) => cons(mark(X), Y) mark(s(X)) => s(mark(X)) a!6220!6220f(X) => f(X) a!6220!6220p(X) => p(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!6220f(0) >? cons(0, f(s(0))) a!6220!6220f(s(0)) >? a!6220!6220f(a!6220!6220p(s(0))) a!6220!6220p(s(X)) >? mark(X) mark(f(X)) >? a!6220!6220f(mark(X)) mark(p(X)) >? a!6220!6220p(mark(X)) mark(0) >? 0 mark(cons(X, Y)) >? cons(mark(X), Y) mark(s(X)) >? s(mark(X)) a!6220!6220f(X) >? f(X) a!6220!6220p(X) >? p(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220f = \y0.2 + y0 a!6220!6220p = \y0.2y0 cons = \y0y1.y0 + y1 f = \y0.1 + y0 mark = \y0.2y0 p = \y0.2y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[a!6220!6220f(0)]] = 2 > 1 = [[cons(0, f(s(0)))]] [[a!6220!6220f(s(0))]] = 2 >= 2 = [[a!6220!6220f(a!6220!6220p(s(0)))]] [[a!6220!6220p(s(_x0))]] = 2x0 >= 2x0 = [[mark(_x0)]] [[mark(f(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[a!6220!6220f(mark(_x0))]] [[mark(p(_x0))]] = 4x0 >= 4x0 = [[a!6220!6220p(mark(_x0))]] [[mark(0)]] = 0 >= 0 = [[0]] [[mark(cons(_x0, _x1))]] = 2x0 + 2x1 >= x1 + 2x0 = [[cons(mark(_x0), _x1)]] [[mark(s(_x0))]] = 2x0 >= 2x0 = [[s(mark(_x0))]] [[a!6220!6220f(_x0)]] = 2 + x0 > 1 + x0 = [[f(_x0)]] [[a!6220!6220p(_x0)]] = 2x0 >= 2x0 = [[p(_x0)]] We can thus remove the following rules: a!6220!6220f(0) => cons(0, f(s(0))) a!6220!6220f(X) => f(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!6220f(s(0)) >? a!6220!6220f(a!6220!6220p(s(0))) a!6220!6220p(s(X)) >? mark(X) mark(f(X)) >? a!6220!6220f(mark(X)) mark(p(X)) >? a!6220!6220p(mark(X)) mark(0) >? 0 mark(cons(X, Y)) >? cons(mark(X), Y) mark(s(X)) >? s(mark(X)) a!6220!6220p(X) >? p(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220f = \y0.y0 a!6220!6220p = \y0.y0 cons = \y0y1.y0 + y1 f = \y0.3 + 3y0 mark = \y0.y0
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