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TRS Stand 20472 pair #381711088
details
property
value
status
complete
benchmark
PEANO_nosorts_noand_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n078.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.671208143234 seconds
cpu usage
0.667855893
max memory
1.0162176E7
stage attributes
key
value
output-size
30120
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 active : [o] --> o mark : [o] --> o plus : [o * o] --> o s : [o] --> o tt : [] --> o active(U11(tt, X, Y)) => mark(U12(tt, X, Y)) active(U12(tt, X, Y)) => mark(s(plus(Y, X))) active(plus(X, 0)) => mark(X) active(plus(X, s(Y))) => mark(U11(tt, Y, X)) mark(U11(X, Y, Z)) => active(U11(mark(X), Y, Z)) mark(tt) => active(tt) mark(U12(X, Y, Z)) => active(U12(mark(X), Y, Z)) mark(s(X)) => active(s(mark(X))) mark(plus(X, Y)) => active(plus(mark(X), mark(Y))) mark(0) => active(0) U11(mark(X), Y, Z) => U11(X, Y, Z) U11(X, mark(Y), Z) => U11(X, Y, Z) U11(X, Y, mark(Z)) => U11(X, Y, Z) U11(active(X), Y, Z) => U11(X, Y, Z) U11(X, active(Y), Z) => U11(X, Y, Z) U11(X, Y, active(Z)) => U11(X, Y, Z) U12(mark(X), Y, Z) => U12(X, Y, Z) U12(X, mark(Y), Z) => U12(X, Y, Z) U12(X, Y, mark(Z)) => U12(X, Y, Z) U12(active(X), Y, Z) => U12(X, Y, Z) U12(X, active(Y), Z) => U12(X, Y, Z) U12(X, Y, active(Z)) => U12(X, Y, Z) s(mark(X)) => s(X) s(active(X)) => s(X) plus(mark(X), Y) => plus(X, Y) plus(X, mark(Y)) => plus(X, Y) plus(active(X), Y) => plus(X, Y) plus(X, active(Y)) => plus(X, Y) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): active(U11(tt, X, Y)) >? mark(U12(tt, X, Y)) active(U12(tt, X, Y)) >? mark(s(plus(Y, X))) active(plus(X, 0)) >? mark(X) active(plus(X, s(Y))) >? mark(U11(tt, Y, X)) mark(U11(X, Y, Z)) >? active(U11(mark(X), Y, Z)) mark(tt) >? active(tt) mark(U12(X, Y, Z)) >? active(U12(mark(X), Y, Z)) mark(s(X)) >? active(s(mark(X))) mark(plus(X, Y)) >? active(plus(mark(X), mark(Y))) mark(0) >? active(0) U11(mark(X), Y, Z) >? U11(X, Y, Z) U11(X, mark(Y), Z) >? U11(X, Y, Z) U11(X, Y, mark(Z)) >? U11(X, Y, Z) U11(active(X), Y, Z) >? U11(X, Y, Z) U11(X, active(Y), Z) >? U11(X, Y, Z) U11(X, Y, active(Z)) >? U11(X, Y, Z) U12(mark(X), Y, Z) >? U12(X, Y, Z) U12(X, mark(Y), Z) >? U12(X, Y, Z) U12(X, Y, mark(Z)) >? U12(X, Y, Z) U12(active(X), Y, Z) >? U12(X, Y, Z) U12(X, active(Y), Z) >? U12(X, Y, Z) U12(X, Y, active(Z)) >? U12(X, Y, Z) s(mark(X)) >? s(X) s(active(X)) >? s(X) plus(mark(X), Y) >? plus(X, Y) plus(X, mark(Y)) >? plus(X, Y) plus(active(X), Y) >? plus(X, Y) plus(X, active(Y)) >? plus(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 U11 = \y0y1y2.2y0 + 2y1 + 2y2 U12 = \y0y1y2.y0 + 2y1 + 2y2 active = \y0.y0 mark = \y0.y0 plus = \y0y1.2y0 + 2y1 s = \y0.y0 tt = 0 Using this interpretation, the requirements translate to: [[active(U11(tt, _x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[mark(U12(tt, _x0, _x1))]] [[active(U12(tt, _x0, _x1))]] = 2x0 + 2x1 >= 2x0 + 2x1 = [[mark(s(plus(_x1, _x0)))]] [[active(plus(_x0, 0))]] = 2 + 2x0 > x0 = [[mark(_x0)]]
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