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HRS union beta 16688 pair #381734350
details
property
value
status
complete
benchmark
kop12thesis_ex7.23.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n070.star.cs.uiowa.edu
space
Kop_13
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.0668659210205 seconds
cpu usage
0.042667645
max memory
1896448.0
stage attributes
key
value
output-size
3168
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: 0 : [] --> o either : [o * o] --> o f : [o -> o * o * o] --> o g : [o * o] --> o s : [o] --> o Rules: f(h, x, 0) => 0 f(h, x, s(y)) => g(y, either(y, h x)) g(x, y) => f(/\z.s(0), y, x) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 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(F, X, 0) >? 0 f(F, X, s(Y)) >? g(Y, either(Y, F X)) g(X, Y) >? f(/\x.s(0), Y, X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 either = \y0y1.y0 + y1 f = \G0y1y2.1 + y1 + 2y2 + G0(y1) g = \y0y1.1 + y1 + 2y0 s = \y0.2y0 Using this interpretation, the requirements translate to: [[f(_F0, _x1, 0)]] = 1 + x1 + F0(x1) > 0 = [[0]] [[f(_F0, _x1, s(_x2))]] = 1 + x1 + 4x2 + F0(x1) >= 1 + x1 + 3x2 + F0(x1) = [[g(_x2, either(_x2, _F0 _x1))]] [[g(_x0, _x1)]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[f(/\x.s(0), _x1, _x0)]] We can thus remove the following rules: f(F, X, 0) => 0 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(F, X, s(Y)) >? g(Y, either(Y, F X)) g(X, Y) >? f(/\x.s(0), Y, X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 either = \y0y1.y0 + y1 f = \G0y1y2.2y1 + 3y2 + 2G0(y1) g = \y0y1.2 + 2y1 + 3y0 s = \y0.1 + 3y0 Using this interpretation, the requirements translate to: [[f(_F0, _x1, s(_x2))]] = 3 + 2x1 + 9x2 + 2F0(x1) > 2 + 2x1 + 5x2 + 2F0(x1) = [[g(_x2, either(_x2, _F0 _x1))]] [[g(_x0, _x1)]] = 2 + 2x1 + 3x0 >= 2 + 2x1 + 3x0 = [[f(/\x.s(0), _x1, _x0)]] We can thus remove the following rules: f(F, X, s(Y)) => g(Y, either(Y, 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]): g(X, Y) >? f(/\x.s(0), Y, X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 f = \G0y1y2.y1 + y2 + G0(0) g = \y0y1.3 + 3y0 + 3y1 s = \y0.y0 Using this interpretation, the requirements translate to: [[g(_x0, _x1)]] = 3 + 3x0 + 3x1 > x0 + x1 = [[f(/\x.s(0), _x1, _x0)]] We can thus remove the following rules:
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