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TRS Stand 20472 pair #381716755
details
property
value
status
complete
benchmark
PEANO_nosorts_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n007.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.274736881256 seconds
cpu usage
0.271263349
max memory
5836800.0
stage attributes
key
value
output-size
16715
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 active : [o] --> o and : [o * o] --> o mark : [o] --> o plus : [o * o] --> o s : [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))) mark(and(X, Y)) => active(and(mark(X), Y)) mark(tt) => active(tt) mark(plus(X, Y)) => active(plus(mark(X), mark(Y))) mark(0) => active(0) mark(s(X)) => active(s(mark(X))) and(mark(X), Y) => and(X, Y) and(X, mark(Y)) => and(X, Y) and(active(X), Y) => and(X, Y) and(X, active(Y)) => and(X, Y) 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) s(mark(X)) => s(X) s(active(X)) => s(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))) mark(and(X, Y)) >? active(and(mark(X), Y)) mark(tt) >? active(tt) mark(plus(X, Y)) >? active(plus(mark(X), mark(Y))) mark(0) >? active(0) mark(s(X)) >? active(s(mark(X))) and(mark(X), Y) >? and(X, Y) and(X, mark(Y)) >? and(X, Y) and(active(X), Y) >? and(X, Y) and(X, active(Y)) >? and(X, Y) 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) s(mark(X)) >? s(X) s(active(X)) >? s(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.y0 + y1 mark = \y0.y0 plus = \y0y1.y1 + 2y0 s = \y0.y0 tt = 2 Using this interpretation, the requirements translate to: [[active(and(tt, _x0))]] = 2 + 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)))]] [[mark(and(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[active(and(mark(_x0), _x1))]] [[mark(tt)]] = 2 >= 2 = [[active(tt)]] [[mark(plus(_x0, _x1))]] = x1 + 2x0 >= x1 + 2x0 = [[active(plus(mark(_x0), mark(_x1)))]] [[mark(0)]] = 0 >= 0 = [[active(0)]] [[mark(s(_x0))]] = x0 >= x0 = [[active(s(mark(_x0)))]] [[and(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[and(_x0, _x1)]] [[and(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[and(_x0, _x1)]] [[and(active(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[and(_x0, _x1)]] [[and(_x0, active(_x1))]] = x0 + x1 >= x0 + x1 = [[and(_x0, _x1)]] [[plus(mark(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[plus(_x0, _x1)]] [[plus(_x0, mark(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[plus(_x0, _x1)]] [[plus(active(_x0), _x1)]] = x1 + 2x0 >= x1 + 2x0 = [[plus(_x0, _x1)]] [[plus(_x0, active(_x1))]] = x1 + 2x0 >= x1 + 2x0 = [[plus(_x0, _x1)]] [[s(mark(_x0))]] = x0 >= x0 = [[s(_x0)]] [[s(active(_x0))]] = x0 >= x0 = [[s(_x0)]] We can thus remove the following rules: active(and(tt, X)) => mark(X) We use rule removal, following [Kop12, Theorem 2.23].
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