details

property	value
status	complete
benchmark	Ex4_7_77_Bor03_C.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n071.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	1.024559021 seconds
cpu usage	0.245398642
max memory	6975488.0

stage attributes

key	value
output-size	15106
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 active : [o] --> o cons : [o * o] --> o mark : [o] --> o ok : [o] --> o proper : [o] --> o tail : [o] --> o top : [o] --> o zeros : [] --> o active(zeros) => mark(cons(0, zeros)) active(tail(cons(X, Y))) => mark(Y) active(cons(X, Y)) => cons(active(X), Y) active(tail(X)) => tail(active(X)) cons(mark(X), Y) => mark(cons(X, Y)) tail(mark(X)) => mark(tail(X)) proper(zeros) => ok(zeros) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(0) => ok(0) proper(tail(X)) => tail(proper(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) tail(ok(X)) => ok(tail(X)) 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(zeros) >? mark(cons(0, zeros)) active(tail(cons(X, Y))) >? mark(Y) active(cons(X, Y)) >? cons(active(X), Y) active(tail(X)) >? tail(active(X)) cons(mark(X), Y) >? mark(cons(X, Y)) tail(mark(X)) >? mark(tail(X)) proper(zeros) >? ok(zeros) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(0) >? ok(0) proper(tail(X)) >? tail(proper(X)) cons(ok(X), ok(Y)) >? ok(cons(X, Y)) tail(ok(X)) >? ok(tail(X)) 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 active = \y0.y0 cons = \y0y1.y1 + 2y0 mark = \y0.y0 ok = \y0.y0 proper = \y0.y0 tail = \y0.2 + y0 top = \y0.y0 zeros = 0 Using this interpretation, the requirements translate to: [[active(zeros)]] = 0 >= 0 = [[mark(cons(0, zeros))]] [[active(tail(cons(_x0, _x1)))]] = 2 + x1 + 2x0 > x1 = [[mark(_x1)]] [[active(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(active(_x0), _x1)]] [[active(tail(_x0))]] = 2 + x0 >= 2 + x0 = [[tail(active(_x0))]] [[cons(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[mark(cons(_x0, _x1))]] [[tail(mark(_x0))]] = 2 + x0 >= 2 + x0 = [[mark(tail(_x0))]] [[proper(zeros)]] = 0 >= 0 = [[ok(zeros)]] [[proper(cons(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[cons(proper(_x0), proper(_x1))]] [[proper(0)]] = 0 >= 0 = [[ok(0)]] [[proper(tail(_x0))]] = 2 + x0 >= 2 + x0 = [[tail(proper(_x0))]] [[cons(ok(_x0), ok(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[ok(cons(_x0, _x1))]] [[tail(ok(_x0))]] = 2 + x0 >= 2 + x0 = [[ok(tail(_x0))]] [[top(mark(_x0))]] = x0 >= x0 = [[top(proper(_x0))]] [[top(ok(_x0))]] = x0 >= x0 = [[top(active(_x0))]] We can thus remove the following rules: active(tail(cons(X, Y))) => mark(Y) We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] active#(zeros) =#> cons#(0, zeros) 1] active#(cons(X, Y)) =#> cons#(active(X), Y) 2] active#(cons(X, Y)) =#> active#(X) 3] active#(tail(X)) =#> tail#(active(X)) popout

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