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TRS Stand 20472 pair #381715985
details
property
value
status
complete
benchmark
Ex6_Luc98_C.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n102.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.43879699707 seconds
cpu usage
0.422439556
max memory
1.359872E7
stage attributes
key
value
output-size
22222
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 cons : [o * o] --> o first : [o * o] --> o from : [o] --> o mark : [o] --> o nil : [] --> o ok : [o] --> o proper : [o] --> o s : [o] --> o top : [o] --> o active(first(0, X)) => mark(nil) active(first(s(X), cons(Y, Z))) => mark(cons(Y, first(X, Z))) active(from(X)) => mark(cons(X, from(s(X)))) active(first(X, Y)) => first(active(X), Y) active(first(X, Y)) => first(X, active(Y)) active(s(X)) => s(active(X)) active(cons(X, Y)) => cons(active(X), Y) active(from(X)) => from(active(X)) first(mark(X), Y) => mark(first(X, Y)) first(X, mark(Y)) => mark(first(X, Y)) s(mark(X)) => mark(s(X)) cons(mark(X), Y) => mark(cons(X, Y)) from(mark(X)) => mark(from(X)) proper(first(X, Y)) => first(proper(X), proper(Y)) proper(0) => ok(0) proper(nil) => ok(nil) proper(s(X)) => s(proper(X)) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(from(X)) => from(proper(X)) first(ok(X), ok(Y)) => ok(first(X, Y)) s(ok(X)) => ok(s(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) from(ok(X)) => ok(from(X)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(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] active#(first(s(X), cons(Y, Z))) =#> cons#(Y, first(X, Z)) 1] active#(first(s(X), cons(Y, Z))) =#> first#(X, Z) 2] active#(from(X)) =#> cons#(X, from(s(X))) 3] active#(from(X)) =#> from#(s(X)) 4] active#(from(X)) =#> s#(X) 5] active#(first(X, Y)) =#> first#(active(X), Y) 6] active#(first(X, Y)) =#> active#(X) 7] active#(first(X, Y)) =#> first#(X, active(Y)) 8] active#(first(X, Y)) =#> active#(Y) 9] active#(s(X)) =#> s#(active(X)) 10] active#(s(X)) =#> active#(X) 11] active#(cons(X, Y)) =#> cons#(active(X), Y) 12] active#(cons(X, Y)) =#> active#(X) 13] active#(from(X)) =#> from#(active(X)) 14] active#(from(X)) =#> active#(X) 15] first#(mark(X), Y) =#> first#(X, Y) 16] first#(X, mark(Y)) =#> first#(X, Y) 17] s#(mark(X)) =#> s#(X) 18] cons#(mark(X), Y) =#> cons#(X, Y) 19] from#(mark(X)) =#> from#(X) 20] proper#(first(X, Y)) =#> first#(proper(X), proper(Y)) 21] proper#(first(X, Y)) =#> proper#(X) 22] proper#(first(X, Y)) =#> proper#(Y) 23] proper#(s(X)) =#> s#(proper(X)) 24] proper#(s(X)) =#> proper#(X) 25] proper#(cons(X, Y)) =#> cons#(proper(X), proper(Y)) 26] proper#(cons(X, Y)) =#> proper#(X) 27] proper#(cons(X, Y)) =#> proper#(Y) 28] proper#(from(X)) =#> from#(proper(X)) 29] proper#(from(X)) =#> proper#(X) 30] first#(ok(X), ok(Y)) =#> first#(X, Y) 31] s#(ok(X)) =#> s#(X) 32] cons#(ok(X), ok(Y)) =#> cons#(X, Y) 33] from#(ok(X)) =#> from#(X) 34] top#(mark(X)) =#> top#(proper(X)) 35] top#(mark(X)) =#> proper#(X) 36] top#(ok(X)) =#> top#(active(X)) 37] top#(ok(X)) =#> active#(X) Rules R_0: active(first(0, X)) => mark(nil) active(first(s(X), cons(Y, Z))) => mark(cons(Y, first(X, Z))) active(from(X)) => mark(cons(X, from(s(X)))) active(first(X, Y)) => first(active(X), Y)
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