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TRS Stand 20472 pair #381713317
details
property
value
status
complete
benchmark
Ex4_7_77_Bor03_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n085.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.187506198883 seconds
cpu usage
0.171728036
max memory
4685824.0
stage attributes
key
value
output-size
12207
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 tail : [o] --> o zeros : [] --> o active(zeros) => mark(cons(0, zeros)) active(tail(cons(X, Y))) => mark(Y) mark(zeros) => active(zeros) mark(cons(X, Y)) => active(cons(mark(X), Y)) mark(0) => active(0) mark(tail(X)) => active(tail(mark(X))) cons(mark(X), Y) => cons(X, Y) cons(X, mark(Y)) => cons(X, Y) cons(active(X), Y) => cons(X, Y) cons(X, active(Y)) => cons(X, Y) tail(mark(X)) => tail(X) tail(active(X)) => tail(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) mark(zeros) >? active(zeros) mark(cons(X, Y)) >? active(cons(mark(X), Y)) mark(0) >? active(0) mark(tail(X)) >? active(tail(mark(X))) cons(mark(X), Y) >? cons(X, Y) cons(X, mark(Y)) >? cons(X, Y) cons(active(X), Y) >? cons(X, Y) cons(X, active(Y)) >? cons(X, Y) tail(mark(X)) >? tail(X) tail(active(X)) >? tail(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.y0 + y1 mark = \y0.y0 tail = \y0.1 + 2y0 zeros = 0 Using this interpretation, the requirements translate to: [[active(zeros)]] = 0 >= 0 = [[mark(cons(0, zeros))]] [[active(tail(cons(_x0, _x1)))]] = 1 + 2x0 + 2x1 > x1 = [[mark(_x1)]] [[mark(zeros)]] = 0 >= 0 = [[active(zeros)]] [[mark(cons(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[active(cons(mark(_x0), _x1))]] [[mark(0)]] = 0 >= 0 = [[active(0)]] [[mark(tail(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[active(tail(mark(_x0)))]] [[cons(mark(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] [[cons(_x0, mark(_x1))]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] [[cons(active(_x0), _x1)]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] [[cons(_x0, active(_x1))]] = x0 + x1 >= x0 + x1 = [[cons(_x0, _x1)]] [[tail(mark(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[tail(_x0)]] [[tail(active(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[tail(_x0)]] We can thus remove the following rules: active(tail(cons(X, Y))) => mark(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(zeros) >? mark(cons(0, zeros)) mark(zeros) >? active(zeros) mark(cons(X, Y)) >? active(cons(mark(X), Y)) mark(0) >? active(0) mark(tail(X)) >? active(tail(mark(X))) cons(mark(X), Y) >? cons(X, Y) cons(X, mark(Y)) >? cons(X, Y) cons(active(X), Y) >? cons(X, Y) cons(X, active(Y)) >? cons(X, Y) tail(mark(X)) >? tail(X) tail(active(X)) >? tail(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 active = \y0.y0
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