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TRS Stand 20472 pair #381715435
details
property
value
status
complete
benchmark
Ex4_7_77_Bor03_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n093.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0235650539398 seconds
cpu usage
0.019462185
max memory
1519616.0
stage attributes
key
value
output-size
2006
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 activate : [o] --> o cons : [o * o] --> o n!6220!6220zeros : [] --> o tail : [o] --> o zeros : [] --> o zeros => cons(0, n!6220!6220zeros) tail(cons(X, Y)) => activate(Y) zeros => n!6220!6220zeros activate(n!6220!6220zeros) => zeros activate(X) => 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]): zeros >? cons(0, n!6220!6220zeros) tail(cons(X, Y)) >? activate(Y) zeros >? n!6220!6220zeros activate(n!6220!6220zeros) >? zeros activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 activate = \y0.3 + y0 cons = \y0y1.1 + y0 + y1 n!6220!6220zeros = 1 tail = \y0.3 + 3y0 zeros = 3 Using this interpretation, the requirements translate to: [[zeros]] = 3 > 2 = [[cons(0, n!6220!6220zeros)]] [[tail(cons(_x0, _x1))]] = 6 + 3x0 + 3x1 > 3 + x1 = [[activate(_x1)]] [[zeros]] = 3 > 1 = [[n!6220!6220zeros]] [[activate(n!6220!6220zeros)]] = 4 > 3 = [[zeros]] [[activate(_x0)]] = 3 + x0 > x0 = [[_x0]] We can thus remove the following rules: zeros => cons(0, n!6220!6220zeros) tail(cons(X, Y)) => activate(Y) zeros => n!6220!6220zeros activate(n!6220!6220zeros) => zeros activate(X) => 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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