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TRS Stand 20472 pair #381715650
details
property
value
status
complete
benchmark
ExSec11_1_Luc02a_L.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n114.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.298033952713 seconds
cpu usage
0.295982291
max memory
9355264.0
stage attributes
key
value
output-size
15090
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 add : [o * o] --> o cons : [o] --> o dbl : [o] --> o first : [o * o] --> o half : [o] --> o nil : [] --> o recip : [o] --> o s : [o] --> o sqr : [o] --> o terms : [o] --> o terms(X) => cons(recip(sqr(X))) sqr(0) => 0 sqr(s(X)) => s(add(sqr(X), dbl(X))) dbl(0) => 0 dbl(s(X)) => s(s(dbl(X))) add(0, X) => X add(s(X), Y) => s(add(X, Y)) first(0, X) => nil first(s(X), cons(Y)) => cons(Y) half(0) => 0 half(s(0)) => 0 half(s(s(X))) => s(half(X)) half(dbl(X)) => 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]): terms(X) >? cons(recip(sqr(X))) sqr(0) >? 0 sqr(s(X)) >? s(add(sqr(X), dbl(X))) dbl(0) >? 0 dbl(s(X)) >? s(s(dbl(X))) add(0, X) >? X add(s(X), Y) >? s(add(X, Y)) first(0, X) >? nil first(s(X), cons(Y)) >? cons(Y) half(0) >? 0 half(s(0)) >? 0 half(s(s(X))) >? s(half(X)) half(dbl(X)) >? X about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[cons(x_1)]] = x_1 [[nil]] = _|_ We choose Lex = {} and Mul = {add, dbl, first, half, recip, s, sqr, terms}, and the following precedence: first > half > terms > recip > dbl = sqr > add > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: terms(X) > recip(sqr(X)) sqr(_|_) >= _|_ sqr(s(X)) >= s(add(sqr(X), dbl(X))) dbl(_|_) >= _|_ dbl(s(X)) >= s(s(dbl(X))) add(_|_, X) > X add(s(X), Y) > s(add(X, Y)) first(_|_, X) >= _|_ first(s(X), Y) > Y half(_|_) >= _|_ half(s(_|_)) >= _|_ half(s(s(X))) > s(half(X)) half(dbl(X)) > X With these choices, we have: 1] terms(X) > recip(sqr(X)) because [2], by definition 2] terms*(X) >= recip(sqr(X)) because terms > recip and [3], by (Copy) 3] terms*(X) >= sqr(X) because terms > sqr and [4], by (Copy) 4] terms*(X) >= X because [5], by (Select) 5] X >= X by (Meta) 6] sqr(_|_) >= _|_ by (Bot) 7] sqr(s(X)) >= s(add(sqr(X), dbl(X))) because [8], by (Star) 8] sqr*(s(X)) >= s(add(sqr(X), dbl(X))) because sqr > s and [9], by (Copy) 9] sqr*(s(X)) >= add(sqr(X), dbl(X)) because sqr > add, [10] and [14], by (Copy) 10] sqr*(s(X)) >= sqr(X) because sqr in Mul and [11], by (Stat) 11] s(X) > X because [12], by definition 12] s*(X) >= X because [13], by (Select) 13] X >= X by (Meta)
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