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TRS Stand 20472 pair #381715680
details
property
value
status
complete
benchmark
Ex9_BLR02_L.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n074.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0460798740387 seconds
cpu usage
0.042514219
max memory
1826816.0
stage attributes
key
value
output-size
3743
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 cons : [o] --> o filter : [o * o * o] --> o nats : [o] --> o s : [o] --> o sieve : [o] --> o zprimes : [] --> o filter(cons(X), 0, Y) => cons(0) filter(cons(X), s(Y), Z) => cons(X) sieve(cons(0)) => cons(0) sieve(cons(s(X))) => cons(s(X)) nats(X) => cons(X) zprimes => sieve(nats(s(s(0)))) As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: 0 : [] --> db cons : [db] --> fb filter : [fb * db * o] --> fb nats : [db] --> fb s : [db] --> db sieve : [fb] --> fb zprimes : [] --> fb We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): filter(cons(X), 0, Y) >? cons(0) filter(cons(X), s(Y), Z) >? cons(X) sieve(cons(0)) >? cons(0) sieve(cons(s(X))) >? cons(s(X)) nats(X) >? cons(X) zprimes >? sieve(nats(s(s(0)))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0.y0 filter = \y0y1y2.3 + y2 + 3y0 + 3y1 nats = \y0.2y0 s = \y0.y0 sieve = \y0.y0 zprimes = 3 Using this interpretation, the requirements translate to: [[filter(cons(_x0), 0, _x1)]] = 3 + x1 + 3x0 > 0 = [[cons(0)]] [[filter(cons(_x0), s(_x1), _x2)]] = 3 + x2 + 3x0 + 3x1 > x0 = [[cons(_x0)]] [[sieve(cons(0))]] = 0 >= 0 = [[cons(0)]] [[sieve(cons(s(_x0)))]] = x0 >= x0 = [[cons(s(_x0))]] [[nats(_x0)]] = 2x0 >= x0 = [[cons(_x0)]] [[zprimes]] = 3 > 0 = [[sieve(nats(s(s(0))))]] We can thus remove the following rules: filter(cons(X), 0, Y) => cons(0) filter(cons(X), s(Y), Z) => cons(X) zprimes => sieve(nats(s(s(0)))) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): sieve(cons(0)) >? cons(0) sieve(cons(s(X))) >? cons(s(X)) nats(X) >? cons(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0.y0 nats = \y0.3 + 3y0 s = \y0.y0 sieve = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[sieve(cons(0))]] = 3 > 0 = [[cons(0)]] [[sieve(cons(s(_x0)))]] = 3 + 3x0 > x0 = [[cons(s(_x0))]] [[nats(_x0)]] = 3 + 3x0 > x0 = [[cons(_x0)]] We can thus remove the following rules:
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