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TRS Stand 20472 pair #381713244
details
property
value
status
complete
benchmark
quick.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n022.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.237153053284 seconds
cpu usage
0.232745593
max memory
9875456.0
stage attributes
key
value
output-size
13270
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 app : [o * o] --> o cons : [o * o] --> o false : [] --> o high : [o * o] --> o ifhigh : [o * o * o] --> o iflow : [o * o * o] --> o le : [o * o] --> o low : [o * o] --> o nil : [] --> o quicksort : [o] --> o s : [o] --> o true : [] --> o le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) low(X, nil) => nil low(X, cons(Y, Z)) => iflow(le(Y, X), X, cons(Y, Z)) iflow(true, X, cons(Y, Z)) => cons(Y, low(X, Z)) iflow(false, X, cons(Y, Z)) => low(X, Z) high(X, nil) => nil high(X, cons(Y, Z)) => ifhigh(le(Y, X), X, cons(Y, Z)) ifhigh(true, X, cons(Y, Z)) => high(X, Z) ifhigh(false, X, cons(Y, Z)) => cons(Y, high(X, Z)) quicksort(nil) => nil quicksort(cons(X, Y)) => app(quicksort(low(X, Y)), cons(X, quicksort(high(X, 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 : [] --> nf app : [nf * nf] --> nf cons : [nf * nf] --> nf false : [] --> rd high : [nf * nf] --> nf ifhigh : [rd * nf * nf] --> nf iflow : [rd * nf * nf] --> nf le : [nf * nf] --> rd low : [nf * nf] --> nf nil : [] --> nf quicksort : [nf] --> nf s : [nf] --> nf true : [] --> rd 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] app#(cons(X, Y), Z) =#> app#(Y, Z) 2] low#(X, cons(Y, Z)) =#> iflow#(le(Y, X), X, cons(Y, Z)) 3] low#(X, cons(Y, Z)) =#> le#(Y, X) 4] iflow#(true, X, cons(Y, Z)) =#> low#(X, Z) 5] iflow#(false, X, cons(Y, Z)) =#> low#(X, Z) 6] high#(X, cons(Y, Z)) =#> ifhigh#(le(Y, X), X, cons(Y, Z)) 7] high#(X, cons(Y, Z)) =#> le#(Y, X) 8] ifhigh#(true, X, cons(Y, Z)) =#> high#(X, Z) 9] ifhigh#(false, X, cons(Y, Z)) =#> high#(X, Z) 10] quicksort#(cons(X, Y)) =#> app#(quicksort(low(X, Y)), cons(X, quicksort(high(X, Y)))) 11] quicksort#(cons(X, Y)) =#> quicksort#(low(X, Y)) 12] quicksort#(cons(X, Y)) =#> low#(X, Y) 13] quicksort#(cons(X, Y)) =#> quicksort#(high(X, Y)) 14] quicksort#(cons(X, Y)) =#> high#(X, Y) Rules R_0: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) low(X, nil) => nil low(X, cons(Y, Z)) => iflow(le(Y, X), X, cons(Y, Z)) iflow(true, X, cons(Y, Z)) => cons(Y, low(X, Z)) iflow(false, X, cons(Y, Z)) => low(X, Z) high(X, nil) => nil high(X, cons(Y, Z)) => ifhigh(le(Y, X), X, cons(Y, Z)) ifhigh(true, X, cons(Y, Z)) => high(X, Z) ifhigh(false, X, cons(Y, Z)) => cons(Y, high(X, Z)) quicksort(nil) => nil quicksort(cons(X, Y)) => app(quicksort(low(X, Y)), cons(X, quicksort(high(X, 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).
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