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TRS Stand 20472 pair #381715885
details
property
value
status
complete
benchmark
Ex9_Luc06_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n055.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.226798057556 seconds
cpu usage
0.222218679
max memory
7106560.0
stage attributes
key
value
output-size
11609
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. a : [] --> o active : [o] --> o b : [] --> o f : [o * o * o] --> o mark : [o] --> o active(f(a, X, X)) => mark(f(X, b, b)) active(b) => mark(a) mark(f(X, Y, Z)) => active(f(X, mark(Y), Z)) mark(a) => active(a) mark(b) => active(b) f(mark(X), Y, Z) => f(X, Y, Z) f(X, mark(Y), Z) => f(X, Y, Z) f(X, Y, mark(Z)) => f(X, Y, Z) f(active(X), Y, Z) => f(X, Y, Z) f(X, active(Y), Z) => f(X, Y, Z) f(X, Y, active(Z)) => f(X, Y, Z) 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, all): Dependency Pairs P_0: 0] active#(f(a, X, X)) =#> mark#(f(X, b, b)) 1] active#(f(a, X, X)) =#> f#(X, b, b) 2] active#(b) =#> mark#(a) 3] mark#(f(X, Y, Z)) =#> active#(f(X, mark(Y), Z)) 4] mark#(f(X, Y, Z)) =#> f#(X, mark(Y), Z) 5] mark#(f(X, Y, Z)) =#> mark#(Y) 6] mark#(a) =#> active#(a) 7] mark#(b) =#> active#(b) 8] f#(mark(X), Y, Z) =#> f#(X, Y, Z) 9] f#(X, mark(Y), Z) =#> f#(X, Y, Z) 10] f#(X, Y, mark(Z)) =#> f#(X, Y, Z) 11] f#(active(X), Y, Z) =#> f#(X, Y, Z) 12] f#(X, active(Y), Z) =#> f#(X, Y, Z) 13] f#(X, Y, active(Z)) =#> f#(X, Y, Z) Rules R_0: active(f(a, X, X)) => mark(f(X, b, b)) active(b) => mark(a) mark(f(X, Y, Z)) => active(f(X, mark(Y), Z)) mark(a) => active(a) mark(b) => active(b) f(mark(X), Y, Z) => f(X, Y, Z) f(X, mark(Y), Z) => f(X, Y, Z) f(X, Y, mark(Z)) => f(X, Y, Z) f(active(X), Y, Z) => f(X, Y, Z) f(X, active(Y), Z) => f(X, Y, Z) f(X, Y, active(Z)) => f(X, Y, Z) Thus, the original system is terminating if (P_0, R_0, minimal, all) is finite. We consider the dependency pair problem (P_0, R_0, minimal, all). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 3, 4, 5 * 1 : 8, 11 * 2 : 6 * 3 : 0, 1 * 4 : 8, 9, 10, 11, 12, 13 * 5 : 3, 4, 5, 6, 7 * 6 : * 7 : 2 * 8 : 8, 9, 10, 11, 12, 13 * 9 : 8, 9, 10, 11, 12, 13 * 10 : 8, 9, 10, 11, 12, 13 * 11 : 8, 9, 10, 11, 12, 13 * 12 : 8, 9, 10, 11, 12, 13 * 13 : 8, 9, 10, 11, 12, 13 This graph has the following strongly connected components: P_1: active#(f(a, X, X)) =#> mark#(f(X, b, b)) mark#(f(X, Y, Z)) =#> active#(f(X, mark(Y), Z)) mark#(f(X, Y, Z)) =#> mark#(Y) P_2: f#(mark(X), Y, Z) =#> f#(X, Y, Z) f#(X, mark(Y), Z) =#> f#(X, Y, Z) f#(X, Y, mark(Z)) =#> f#(X, Y, Z) f#(active(X), Y, Z) =#> f#(X, Y, Z) f#(X, active(Y), Z) =#> f#(X, Y, Z)
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