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TRS Stand 20472 pair #381716345
details
property
value
status
complete
benchmark
Ex6_Luc98_Z.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)
0.0496711730957 seconds
cpu usage
0.045264689
max memory
1720320.0
stage attributes
key
value
output-size
4737
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 activate : [o] --> o cons : [o * o] --> o first : [o * o] --> o from : [o] --> o n!6220!6220first : [o * o] --> o n!6220!6220from : [o] --> o nil : [] --> o s : [o] --> o first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) from(X) => cons(X, n!6220!6220from(s(X))) first(X, Y) => n!6220!6220first(X, Y) from(X) => n!6220!6220from(X) activate(n!6220!6220first(X, Y)) => first(X, Y) activate(n!6220!6220from(X)) => from(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]): first(0, X) >? nil first(s(X), cons(Y, Z)) >? cons(Y, n!6220!6220first(X, activate(Z))) from(X) >? cons(X, n!6220!6220from(s(X))) first(X, Y) >? n!6220!6220first(X, Y) from(X) >? n!6220!6220from(X) activate(n!6220!6220first(X, Y)) >? first(X, Y) activate(n!6220!6220from(X)) >? from(X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 activate = \y0.2y0 cons = \y0y1.y0 + y1 first = \y0y1.y0 + 2y1 from = \y0.2 + 2y0 n!6220!6220first = \y0y1.y0 + y1 n!6220!6220from = \y0.1 + y0 nil = 0 s = \y0.1 + y0 Using this interpretation, the requirements translate to: [[first(0, _x0)]] = 3 + 2x0 > 0 = [[nil]] [[first(s(_x0), cons(_x1, _x2))]] = 1 + x0 + 2x1 + 2x2 > x0 + x1 + 2x2 = [[cons(_x1, n!6220!6220first(_x0, activate(_x2)))]] [[from(_x0)]] = 2 + 2x0 >= 2 + 2x0 = [[cons(_x0, n!6220!6220from(s(_x0)))]] [[first(_x0, _x1)]] = x0 + 2x1 >= x0 + x1 = [[n!6220!6220first(_x0, _x1)]] [[from(_x0)]] = 2 + 2x0 > 1 + x0 = [[n!6220!6220from(_x0)]] [[activate(n!6220!6220first(_x0, _x1))]] = 2x0 + 2x1 >= x0 + 2x1 = [[first(_x0, _x1)]] [[activate(n!6220!6220from(_x0))]] = 2 + 2x0 >= 2 + 2x0 = [[from(_x0)]] [[activate(_x0)]] = 2x0 >= x0 = [[_x0]] We can thus remove the following rules: first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) from(X) => n!6220!6220from(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]): from(X) >? cons(X, n!6220!6220from(s(X))) first(X, Y) >? n!6220!6220first(X, Y) activate(n!6220!6220first(X, Y)) >? first(X, Y) activate(n!6220!6220from(X)) >? from(X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: activate = \y0.3 + 3y0 cons = \y0y1.y0 + y1 first = \y0y1.1 + y1 + 2y0 from = \y0.2 + 2y0 n!6220!6220first = \y0y1.1 + y0 + y1 n!6220!6220from = \y0.1 + y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[from(_x0)]] = 2 + 2x0 > 1 + 2x0 = [[cons(_x0, n!6220!6220from(s(_x0)))]] [[first(_x0, _x1)]] = 1 + x1 + 2x0 >= 1 + x0 + x1 = [[n!6220!6220first(_x0, _x1)]]
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