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TRS Stand 20472 pair #381716275
details
property
value
status
complete
benchmark
Ex6_GM04_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n082.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0278689861298 seconds
cpu usage
0.024410051
max memory
1298432.0
stage attributes
key
value
output-size
2181
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. a!6220!6220c : [] --> o a!6220!6220f : [o] --> o c : [] --> o f : [o] --> o g : [o] --> o mark : [o] --> o a!6220!6220c => a!6220!6220f(g(c)) a!6220!6220f(g(X)) => g(X) mark(c) => a!6220!6220c mark(f(X)) => a!6220!6220f(X) mark(g(X)) => g(X) a!6220!6220c => c 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!6220c >? a!6220!6220f(g(c)) a!6220!6220f(g(X)) >? g(X) mark(c) >? a!6220!6220c mark(f(X)) >? a!6220!6220f(X) mark(g(X)) >? g(X) a!6220!6220c >? c a!6220!6220f(X) >? f(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220c = 2 a!6220!6220f = \y0.1 + 2y0 c = 0 f = \y0.y0 g = \y0.2y0 mark = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[a!6220!6220c]] = 2 > 1 = [[a!6220!6220f(g(c))]] [[a!6220!6220f(g(_x0))]] = 1 + 4x0 > 2x0 = [[g(_x0)]] [[mark(c)]] = 3 > 2 = [[a!6220!6220c]] [[mark(f(_x0))]] = 3 + 3x0 > 1 + 2x0 = [[a!6220!6220f(_x0)]] [[mark(g(_x0))]] = 3 + 6x0 > 2x0 = [[g(_x0)]] [[a!6220!6220c]] = 2 > 0 = [[c]] [[a!6220!6220f(_x0)]] = 1 + 2x0 > x0 = [[f(_x0)]] We can thus remove the following rules: a!6220!6220c => a!6220!6220f(g(c)) a!6220!6220f(g(X)) => g(X) mark(c) => a!6220!6220c mark(f(X)) => a!6220!6220f(X) mark(g(X)) => g(X) a!6220!6220c => c a!6220!6220f(X) => f(X) All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.
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