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TRS Stand 20472 pair #381715740
details
property
value
status
complete
benchmark
Ex9_BLR02_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n048.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.229752779007 seconds
cpu usage
0.22618707
max memory
1.0059776E7
stage attributes
key
value
output-size
10147
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 activate : [o] --> o cons : [o * o] --> o filter : [o * o * o] --> o n!6220!6220filter : [o * o * o] --> o n!6220!6220nats : [o] --> o n!6220!6220sieve : [o] --> o nats : [o] --> o s : [o] --> o sieve : [o] --> o zprimes : [] --> o filter(cons(X, Y), 0, Z) => cons(0, n!6220!6220filter(activate(Y), Z, Z)) filter(cons(X, Y), s(Z), U) => cons(X, n!6220!6220filter(activate(Y), Z, U)) sieve(cons(0, X)) => cons(0, n!6220!6220sieve(activate(X))) sieve(cons(s(X), Y)) => cons(s(X), n!6220!6220sieve(filter(activate(Y), X, X))) nats(X) => cons(X, n!6220!6220nats(s(X))) zprimes => sieve(nats(s(s(0)))) filter(X, Y, Z) => n!6220!6220filter(X, Y, Z) sieve(X) => n!6220!6220sieve(X) nats(X) => n!6220!6220nats(X) activate(n!6220!6220filter(X, Y, Z)) => filter(X, Y, Z) activate(n!6220!6220sieve(X)) => sieve(X) activate(n!6220!6220nats(X)) => nats(X) activate(X) => X 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] filter#(cons(X, Y), 0, Z) =#> activate#(Y) 1] filter#(cons(X, Y), s(Z), U) =#> activate#(Y) 2] sieve#(cons(0, X)) =#> activate#(X) 3] sieve#(cons(s(X), Y)) =#> filter#(activate(Y), X, X) 4] sieve#(cons(s(X), Y)) =#> activate#(Y) 5] zprimes# =#> sieve#(nats(s(s(0)))) 6] zprimes# =#> nats#(s(s(0))) 7] activate#(n!6220!6220filter(X, Y, Z)) =#> filter#(X, Y, Z) 8] activate#(n!6220!6220sieve(X)) =#> sieve#(X) 9] activate#(n!6220!6220nats(X)) =#> nats#(X) Rules R_0: filter(cons(X, Y), 0, Z) => cons(0, n!6220!6220filter(activate(Y), Z, Z)) filter(cons(X, Y), s(Z), U) => cons(X, n!6220!6220filter(activate(Y), Z, U)) sieve(cons(0, X)) => cons(0, n!6220!6220sieve(activate(X))) sieve(cons(s(X), Y)) => cons(s(X), n!6220!6220sieve(filter(activate(Y), X, X))) nats(X) => cons(X, n!6220!6220nats(s(X))) zprimes => sieve(nats(s(s(0)))) filter(X, Y, Z) => n!6220!6220filter(X, Y, Z) sieve(X) => n!6220!6220sieve(X) nats(X) => n!6220!6220nats(X) activate(n!6220!6220filter(X, Y, Z)) => filter(X, Y, Z) activate(n!6220!6220sieve(X)) => sieve(X) activate(n!6220!6220nats(X)) => nats(X) activate(X) => X Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 7, 8, 9 * 1 : 7, 8, 9 * 2 : 7, 8, 9 * 3 : 0, 1 * 4 : 7, 8, 9 * 5 : 2, 3, 4 * 6 : * 7 : 0, 1 * 8 : 2, 3, 4 * 9 : This graph has the following strongly connected components: P_1: filter#(cons(X, Y), 0, Z) =#> activate#(Y) filter#(cons(X, Y), s(Z), U) =#> activate#(Y) sieve#(cons(0, X)) =#> activate#(X) sieve#(cons(s(X), Y)) =#> filter#(activate(Y), X, X) sieve#(cons(s(X), Y)) =#> activate#(Y) activate#(n!6220!6220filter(X, Y, Z)) =#> filter#(X, Y, Z) activate#(n!6220!6220sieve(X)) =#> sieve#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f).
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