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TRS Stand 20472 pair #381716520
details
property
value
status
complete
benchmark
Ex5_Zan97_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n049.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0345990657806 seconds
cpu usage
0.02848621
max memory
1396736.0
stage attributes
key
value
output-size
2786
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. activate : [o] --> o c : [] --> o f : [o] --> o false : [] --> o if : [o * o * o] --> o n!6220!6220f : [o] --> o true : [] --> o f(X) => if(X, c, n!6220!6220f(true)) if(true, X, Y) => X if(false, X, Y) => activate(Y) f(X) => n!6220!6220f(X) activate(n!6220!6220f(X)) => f(X) 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]): f(X) >? if(X, c, n!6220!6220f(true)) if(true, X, Y) >? X if(false, X, Y) >? activate(Y) f(X) >? n!6220!6220f(X) activate(n!6220!6220f(X)) >? f(X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate = \y0.2 + 2y0 c = 0 f = \y0.1 + y0 false = 3 if = \y0y1y2.1 + y0 + y1 + 2y2 n!6220!6220f = \y0.y0 true = 0 Using this interpretation, the requirements translate to: [[f(_x0)]] = 1 + x0 >= 1 + x0 = [[if(_x0, c, n!6220!6220f(true))]] [[if(true, _x0, _x1)]] = 1 + x0 + 2x1 > x0 = [[_x0]] [[if(false, _x0, _x1)]] = 4 + x0 + 2x1 > 2 + 2x1 = [[activate(_x1)]] [[f(_x0)]] = 1 + x0 > x0 = [[n!6220!6220f(_x0)]] [[activate(n!6220!6220f(_x0))]] = 2 + 2x0 > 1 + x0 = [[f(_x0)]] [[activate(_x0)]] = 2 + 2x0 > x0 = [[_x0]] We can thus remove the following rules: if(true, X, Y) => X if(false, X, Y) => activate(Y) f(X) => n!6220!6220f(X) activate(n!6220!6220f(X)) => f(X) 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]): f(X) >? if(X, c, n!6220!6220f(true)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: c = 0 f = \y0.3 + 3y0 if = \y0y1y2.y0 + y1 + y2 n!6220!6220f = \y0.y0 true = 0 Using this interpretation, the requirements translate to: [[f(_x0)]] = 3 + 3x0 > x0 = [[if(_x0, c, n!6220!6220f(true))]] We can thus remove the following rules: f(X) => if(X, c, n!6220!6220f(true)) 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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