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TRS Stand 20472 pair #381713243
details
property
value
status
complete
benchmark
perfect.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n079.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.108453989029 seconds
cpu usage
0.089369275
max memory
6262784.0
stage attributes
key
value
output-size
4227
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 f : [o * o * o * o] --> o false : [] --> o if : [o * o * o] --> o le : [o * o] --> o minus : [o * o] --> o perfectp : [o] --> o s : [o] --> o true : [] --> o perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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 : [] --> bc f : [bc * bc * bc * bc] --> kc false : [] --> kc if : [wb * kc * kc] --> kc le : [bc * bc] --> wb minus : [bc * bc] --> bc perfectp : [bc] --> kc s : [bc] --> bc true : [] --> kc 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] perfectp#(s(X)) =#> f#(X, s(0), s(X), s(X)) 1] f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) 2] f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) 3] f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) Rules R_0: perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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 : 2, 3 * 1 : 1, 2, 3 * 2 : * 3 : 1, 2, 3 This graph has the following strongly connected components: P_1: f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) 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). Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(f#) = 1 Thus, we can orient the dependency pairs as follows: nu(f#(s(X), 0, Y, Z)) = s(X) |> X = nu(f#(X, Z, minus(Y, s(X)), Z)) nu(f#(s(X), s(Y), Z, U)) = s(X) |> X = nu(f#(X, U, Z, U)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++
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