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TRS Stand 20472 pair #381709909
details
property
value
status
complete
benchmark
division.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n035.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.126888036728 seconds
cpu usage
0.117357649
max memory
4968448.0
stage attributes
key
value
output-size
7800
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 false : [] --> o ifMinus : [o * o * o] --> o le : [o * o] --> o minus : [o * o] --> o quot : [o * o] --> o s : [o] --> o true : [] --> o le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(0, X) => 0 minus(s(X), Y) => ifMinus(le(s(X), Y), s(X), Y) ifMinus(true, s(X), Y) => 0 ifMinus(false, s(X), Y) => s(minus(X, Y)) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(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 : [] --> nc false : [] --> ya ifMinus : [ya * nc * nc] --> nc le : [nc * nc] --> ya minus : [nc * nc] --> nc quot : [nc * nc] --> nc s : [nc] --> nc true : [] --> ya 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] le#(s(X), s(Y)) =#> le#(X, Y) 1] minus#(s(X), Y) =#> ifMinus#(le(s(X), Y), s(X), Y) 2] minus#(s(X), Y) =#> le#(s(X), Y) 3] ifMinus#(false, s(X), Y) =#> minus#(X, Y) 4] quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(Y)) 5] quot#(s(X), s(Y)) =#> minus#(X, Y) Rules R_0: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(0, X) => 0 minus(s(X), Y) => ifMinus(le(s(X), Y), s(X), Y) ifMinus(true, s(X), Y) => 0 ifMinus(false, s(X), Y) => s(minus(X, Y)) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(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 : 3 * 2 : 0 * 3 : 1, 2 * 4 : 4, 5 * 5 : 1, 2 This graph has the following strongly connected components: P_1: le#(s(X), s(Y)) =#> le#(X, Y) P_2: minus#(s(X), Y) =#> ifMinus#(le(s(X), Y), s(X), Y) ifMinus#(false, s(X), Y) =#> minus#(X, Y) P_3: quot#(s(X), s(Y)) =#> quot#(minus(X, Y), s(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), (P_2, R_0, m, f) and (P_3, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative).
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