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TRS Stand 20472 pair #381715230
details
property
value
status
complete
benchmark
Ex2_Luc02a_L.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
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.141125917435 seconds
cpu usage
0.137877501
max memory
5689344.0
stage attributes
key
value
output-size
7806
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 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) 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) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[nil]] = _|_ We choose Lex = {} and Mul = {add, cons, dbl, first, recip, s, sqr, terms}, and the following precedence: terms > cons = first > 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) >= cons(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), cons(Y)) >= cons(Y) With these choices, we have: 1] terms(X) >= cons(recip(sqr(X))) because [2], by (Star) 2] terms*(X) >= cons(recip(sqr(X))) because terms > cons and [3], by (Copy) 3] terms*(X) >= recip(sqr(X)) because terms > recip and [4], by (Copy) 4] terms*(X) >= sqr(X) because terms > sqr and [5], by (Copy) 5] terms*(X) >= X because [6], by (Select) 6] X >= X by (Meta) 7] sqr(_|_) >= _|_ by (Bot) 8] sqr(s(X)) >= s(add(sqr(X), dbl(X))) because [9], by (Star) 9] sqr*(s(X)) >= s(add(sqr(X), dbl(X))) because sqr > s and [10], by (Copy) 10] sqr*(s(X)) >= add(sqr(X), dbl(X)) because sqr > add, [11] and [15], by (Copy) 11] sqr*(s(X)) >= sqr(X) because sqr in Mul and [12], by (Stat) 12] s(X) > X because [13], by definition 13] s*(X) >= X because [14], by (Select) 14] X >= X by (Meta) 15] sqr*(s(X)) >= dbl(X) because sqr = dbl, sqr in Mul and [12], by (Stat) 16] dbl(_|_) >= _|_ by (Bot) 17] dbl(s(X)) >= s(s(dbl(X))) because [18], by (Star) 18] dbl*(s(X)) >= s(s(dbl(X))) because dbl > s and [19], by (Copy) 19] dbl*(s(X)) >= s(dbl(X)) because dbl > s and [20], by (Copy) 20] dbl*(s(X)) >= dbl(X) because dbl in Mul and [12], by (Stat) 21] add(_|_, X) >= X because [22], by (Star) 22] add*(_|_, X) >= X because [14], by (Select) 23] add(s(X), Y) > s(add(X, Y)) because [24], by definition
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