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TRS Stand 20472 pair #381716400
details
property
value
status
complete
benchmark
Ex1_Zan97_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n002.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.186438083649 seconds
cpu usage
0.181784072
max memory
3104768.0
stage attributes
key
value
output-size
6194
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. active : [o] --> o c : [] --> o d : [] --> o g : [o] --> o h : [o] --> o mark : [o] --> o active(g(X)) => mark(h(X)) active(c) => mark(d) active(h(d)) => mark(g(c)) mark(g(X)) => active(g(X)) mark(h(X)) => active(h(X)) mark(c) => active(c) mark(d) => active(d) g(mark(X)) => g(X) g(active(X)) => g(X) h(mark(X)) => h(X) h(active(X)) => h(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(g(X)) >? mark(h(X)) active(c) >? mark(d) active(h(d)) >? mark(g(c)) mark(g(X)) >? active(g(X)) mark(h(X)) >? active(h(X)) mark(c) >? active(c) mark(d) >? active(d) g(mark(X)) >? g(X) g(active(X)) >? g(X) h(mark(X)) >? h(X) h(active(X)) >? h(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: active = \y0.1 + y0 c = 0 d = 0 g = \y0.2y0 h = \y0.y0 mark = \y0.1 + 2y0 Using this interpretation, the requirements translate to: [[active(g(_x0))]] = 1 + 2x0 >= 1 + 2x0 = [[mark(h(_x0))]] [[active(c)]] = 1 >= 1 = [[mark(d)]] [[active(h(d))]] = 1 >= 1 = [[mark(g(c))]] [[mark(g(_x0))]] = 1 + 4x0 >= 1 + 2x0 = [[active(g(_x0))]] [[mark(h(_x0))]] = 1 + 2x0 >= 1 + x0 = [[active(h(_x0))]] [[mark(c)]] = 1 >= 1 = [[active(c)]] [[mark(d)]] = 1 >= 1 = [[active(d)]] [[g(mark(_x0))]] = 2 + 4x0 > 2x0 = [[g(_x0)]] [[g(active(_x0))]] = 2 + 2x0 > 2x0 = [[g(_x0)]] [[h(mark(_x0))]] = 1 + 2x0 > x0 = [[h(_x0)]] [[h(active(_x0))]] = 1 + x0 > x0 = [[h(_x0)]] We can thus remove the following rules: g(mark(X)) => g(X) g(active(X)) => g(X) h(mark(X)) => h(X) h(active(X)) => h(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#(g(X)) =#> mark#(h(X)) 1] active#(c) =#> mark#(d) 2] active#(h(d)) =#> mark#(g(c)) 3] mark#(g(X)) =#> active#(g(X)) 4] mark#(h(X)) =#> active#(h(X)) 5] mark#(c) =#> active#(c) 6] mark#(d) =#> active#(d) Rules R_0: active(g(X)) => mark(h(X)) active(c) => mark(d) active(h(d)) => mark(g(c)) mark(g(X)) => active(g(X)) mark(h(X)) => active(h(X)) mark(c) => active(c)
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