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TRS Stand 20472 pair #381715200
details
property
value
status
complete
benchmark
PEANO_nosorts_C.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n010.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.397056102753 seconds
cpu usage
0.393538652
max memory
8179712.0
stage attributes
key
value
output-size
20636
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 and : [o * o] --> o mark : [o] --> o ok : [o] --> o plus : [o * o] --> o proper : [o] --> o s : [o] --> o top : [o] --> o tt : [] --> o active(and(tt, X)) => mark(X) active(plus(X, 0)) => mark(X) active(plus(X, s(Y))) => mark(s(plus(X, Y))) active(and(X, Y)) => and(active(X), Y) active(plus(X, Y)) => plus(active(X), Y) active(plus(X, Y)) => plus(X, active(Y)) active(s(X)) => s(active(X)) and(mark(X), Y) => mark(and(X, Y)) plus(mark(X), Y) => mark(plus(X, Y)) plus(X, mark(Y)) => mark(plus(X, Y)) s(mark(X)) => mark(s(X)) proper(and(X, Y)) => and(proper(X), proper(Y)) proper(tt) => ok(tt) proper(plus(X, Y)) => plus(proper(X), proper(Y)) proper(0) => ok(0) proper(s(X)) => s(proper(X)) and(ok(X), ok(Y)) => ok(and(X, Y)) plus(ok(X), ok(Y)) => ok(plus(X, Y)) s(ok(X)) => ok(s(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(and(tt, X)) >? mark(X) active(plus(X, 0)) >? mark(X) active(plus(X, s(Y))) >? mark(s(plus(X, Y))) active(and(X, Y)) >? and(active(X), Y) active(plus(X, Y)) >? plus(active(X), Y) active(plus(X, Y)) >? plus(X, active(Y)) active(s(X)) >? s(active(X)) and(mark(X), Y) >? mark(and(X, Y)) plus(mark(X), Y) >? mark(plus(X, Y)) plus(X, mark(Y)) >? mark(plus(X, Y)) s(mark(X)) >? mark(s(X)) proper(and(X, Y)) >? and(proper(X), proper(Y)) proper(tt) >? ok(tt) proper(plus(X, Y)) >? plus(proper(X), proper(Y)) proper(0) >? ok(0) proper(s(X)) >? s(proper(X)) and(ok(X), ok(Y)) >? ok(and(X, Y)) plus(ok(X), ok(Y)) >? ok(plus(X, Y)) s(ok(X)) >? ok(s(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 and = \y0y1.1 + y1 + 2y0 mark = \y0.y0 ok = \y0.y0 plus = \y0y1.y1 + 2y0 proper = \y0.y0 s = \y0.y0 top = \y0.y0 tt = 0 Using this interpretation, the requirements translate to: [[active(and(tt, _x0))]] = 1 + x0 > x0 = [[mark(_x0)]] [[active(plus(_x0, 0))]] = 2x0 >= x0 = [[mark(_x0)]] [[active(plus(_x0, s(_x1)))]] = x1 + 2x0 >= x1 + 2x0 = [[mark(s(plus(_x0, _x1)))]] [[active(and(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[and(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[plus(active(_x0), _x1)]] [[active(plus(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[plus(_x0, active(_x1))]] [[active(s(_x0))]] = x0 >= x0 = [[s(active(_x0))]] [[and(mark(_x0), _x1)]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[mark(and(_x0, _x1))]] [[plus(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[mark(plus(_x0, _x1))]] [[plus(_x0, mark(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[mark(plus(_x0, _x1))]] [[s(mark(_x0))]] = x0 >= x0 = [[mark(s(_x0))]] [[proper(and(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[and(proper(_x0), proper(_x1))]] [[proper(tt)]] = 0 >= 0 = [[ok(tt)]]
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