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TRS Stand 20472 pair #381713311
details
property
value
status
complete
benchmark
#4.34.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n006.star.cs.uiowa.edu
space
Strategy_removed_AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.100521087646 seconds
cpu usage
0.096852149
max memory
4218880.0
stage attributes
key
value
output-size
5860
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. 0 : [] --> o 1 : [] --> o c : [o] --> o f : [o] --> o false : [] --> o g : [o * o] --> o if : [o * o * o] --> o s : [o] --> o true : [] --> o f(0) => true f(1) => false f(s(X)) => f(X) if(true, X, Y) => X if(false, X, Y) => Y g(s(X), s(Y)) => if(f(X), s(X), s(Y)) g(X, c(Y)) => g(X, g(s(c(Y)), Y)) 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 : [] --> pb 1 : [] --> pb c : [pb] --> pb f : [pb] --> xa false : [] --> xa g : [pb * pb] --> pb if : [xa * pb * pb] --> pb s : [pb] --> pb true : [] --> xa 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] f#(s(X)) =#> f#(X) 1] g#(s(X), s(Y)) =#> if#(f(X), s(X), s(Y)) 2] g#(s(X), s(Y)) =#> f#(X) 3] g#(X, c(Y)) =#> g#(X, g(s(c(Y)), Y)) 4] g#(X, c(Y)) =#> g#(s(c(Y)), Y) Rules R_0: f(0) => true f(1) => false f(s(X)) => f(X) if(true, X, Y) => X if(false, X, Y) => Y g(s(X), s(Y)) => if(f(X), s(X), s(Y)) g(X, c(Y)) => g(X, g(s(c(Y)), Y)) 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 : 0 * 1 : * 2 : 0 * 3 : 1, 2, 3, 4 * 4 : 1, 2, 3, 4 This graph has the following strongly connected components: P_1: f#(s(X)) =#> f#(X) P_2: g#(X, c(Y)) =#> g#(X, g(s(c(Y)), Y)) g#(X, c(Y)) =#> g#(s(c(Y)), Y) 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) and (P_2, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: g#(X, c(Y)) >? g#(X, g(s(c(Y)), Y)) g#(X, c(Y)) >? g#(s(c(Y)), Y) f(0) >= true f(1) >= false f(s(X)) >= f(X) if(true, X, Y) >= X
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