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TRS Stand 20472 pair #381714010
details
property
value
status
complete
benchmark
PEANO_nosorts-noand_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n041.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0715310573578 seconds
cpu usage
0.057691436
max memory
1814528.0
stage attributes
key
value
output-size
5217
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 U11 : [o * o * o] --> o U12 : [o * o * o] --> o activate : [o] --> o plus : [o * o] --> o s : [o] --> o tt : [] --> o U11(tt, X, Y) => U12(tt, activate(X), activate(Y)) U12(tt, X, Y) => s(plus(activate(Y), activate(X))) plus(X, 0) => X plus(X, s(Y)) => U11(tt, Y, X) activate(X) => X As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: 0 : [] --> ab U11 : [h * ab * ab] --> ab U12 : [h * ab * ab] --> ab activate : [ab] --> ab plus : [ab * ab] --> ab s : [ab] --> ab tt : [] --> h We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): U11(tt, X, Y) >? U12(tt, activate(X), activate(Y)) U12(tt, X, Y) >? s(plus(activate(Y), activate(X))) plus(X, 0) >? X plus(X, s(Y)) >? U11(tt, Y, X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 U11 = \y0y1y2.y0 + y1 + y2 U12 = \y0y1y2.y0 + y1 + y2 activate = \y0.y0 plus = \y0y1.y0 + y1 s = \y0.y0 tt = 0 Using this interpretation, the requirements translate to: [[U11(tt, _x0, _x1)]] = x0 + x1 >= x0 + x1 = [[U12(tt, activate(_x0), activate(_x1))]] [[U12(tt, _x0, _x1)]] = x0 + x1 >= x0 + x1 = [[s(plus(activate(_x1), activate(_x0)))]] [[plus(_x0, 0)]] = 3 + x0 > x0 = [[_x0]] [[plus(_x0, s(_x1))]] = x0 + x1 >= x0 + x1 = [[U11(tt, _x1, _x0)]] [[activate(_x0)]] = x0 >= x0 = [[_x0]] We can thus remove the following rules: plus(X, 0) => 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]): U11(tt, X, Y) >? U12(tt, activate(X), activate(Y)) U12(tt, X, Y) >? s(plus(activate(Y), activate(X))) plus(X, s(Y)) >? U11(tt, Y, X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: U11 = \y0y1y2.1 + y2 + 2y0 + 3y1 U12 = \y0y1y2.1 + y2 + 2y0 + 3y1 activate = \y0.y0 plus = \y0y1.1 + y0 + 3y1 s = \y0.1 + y0 tt = 1 Using this interpretation, the requirements translate to: [[U11(tt, _x0, _x1)]] = 3 + x1 + 3x0 >= 3 + x1 + 3x0 = [[U12(tt, activate(_x0), activate(_x1))]] [[U12(tt, _x0, _x1)]] = 3 + x1 + 3x0 > 2 + x1 + 3x0 = [[s(plus(activate(_x1), activate(_x0)))]] [[plus(_x0, s(_x1))]] = 4 + x0 + 3x1 > 3 + x0 + 3x1 = [[U11(tt, _x1, _x0)]] [[activate(_x0)]] = x0 >= x0 = [[_x0]] We can thus remove the following rules: U12(tt, X, Y) => s(plus(activate(Y), activate(X))) plus(X, s(Y)) => U11(tt, Y, X)
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