details

property	value
status	complete
benchmark	PEANO_nosorts_noand_C.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n089.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.931583881378 seconds
cpu usage	0.926716636
max memory	1.4036992E7

stage attributes

key	value
output-size	35960
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 U11 : [o * o * o] --> o U12 : [o * o * o] --> o active : [o] --> o mark : [o] --> o ok : [o] --> o plus : [o * o] --> o proper : [o] --> o s : [o] --> o top : [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)) active(U11(X, Y, Z)) => U11(active(X), Y, Z) active(U12(X, Y, Z)) => U12(active(X), Y, Z) active(s(X)) => s(active(X)) active(plus(X, Y)) => plus(active(X), Y) active(plus(X, Y)) => plus(X, active(Y)) U11(mark(X), Y, Z) => mark(U11(X, Y, Z)) U12(mark(X), Y, Z) => mark(U12(X, Y, Z)) s(mark(X)) => mark(s(X)) plus(mark(X), Y) => mark(plus(X, Y)) plus(X, mark(Y)) => mark(plus(X, Y)) proper(U11(X, Y, Z)) => U11(proper(X), proper(Y), proper(Z)) proper(tt) => ok(tt) proper(U12(X, Y, Z)) => U12(proper(X), proper(Y), proper(Z)) proper(s(X)) => s(proper(X)) proper(plus(X, Y)) => plus(proper(X), proper(Y)) proper(0) => ok(0) U11(ok(X), ok(Y), ok(Z)) => ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) => ok(U12(X, Y, Z)) s(ok(X)) => ok(s(X)) plus(ok(X), ok(Y)) => ok(plus(X, Y)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(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]): 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)) active(U11(X, Y, Z)) >? U11(active(X), Y, Z) active(U12(X, Y, Z)) >? U12(active(X), Y, Z) active(s(X)) >? s(active(X)) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) U11(mark(X), Y, Z) >? mark(U11(X, Y, Z)) U12(mark(X), Y, Z) >? mark(U12(X, Y, Z)) s(mark(X)) >? mark(s(X)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) proper(U11(X, Y, Z)) >? U11(proper(X), proper(Y), proper(Z)) proper(tt) >? ok(tt) proper(U12(X, Y, Z)) >? U12(proper(X), proper(Y), proper(Z)) proper(s(X)) >? s(proper(X)) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) U11(ok(X), ok(Y), ok(Z)) >? ok(U11(X, Y, Z)) U12(ok(X), ok(Y), ok(Z)) >? ok(U12(X, Y, Z)) s(ok(X)) >? ok(s(X)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) top(mark(X)) >? top(proper(X)) top(ok(X)) >? top(active(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 U11 = \y0y1y2.2 + y2 + 2y0 + 2y1 U12 = \y0y1y2.2 + y2 + 2y0 + 2y1 active = \y0.y0 mark = \y0.y0 ok = \y0.y0 plus = \y0y1.y0 + 2y1 proper = \y0.y0 s = \y0.2 + y0 top = \y0.2y0 tt = 0 Using this interpretation, the requirements translate to: [[active(U11(tt, _x0, _x1))]] = 2 + x1 + 2x0 >= 2 + x1 + 2x0 = [[mark(U12(tt, _x0, _x1))]] popout

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all output return to TRS Stand 20472