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TRS Stand 20472 pair #381714824
details
property
value
status
complete
benchmark
Ex7_BLR02_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n192.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.171061038971 seconds
cpu usage
0.165892127
max memory
5652480.0
stage attributes
key
value
output-size
11472
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 2nd : [o] --> o activate : [o] --> o cons : [o * o] --> o from : [o] --> o head : [o] --> o n!6220!6220from : [o] --> o n!6220!6220take : [o * o] --> o nil : [] --> o s : [o] --> o sel : [o * o] --> o take : [o * o] --> o from(X) => cons(X, n!6220!6220from(s(X))) head(cons(X, Y)) => X 2nd(cons(X, Y)) => head(activate(Y)) take(0, X) => nil take(s(X), cons(Y, Z)) => cons(Y, n!6220!6220take(X, activate(Z))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) from(X) => n!6220!6220from(X) take(X, Y) => n!6220!6220take(X, Y) activate(n!6220!6220from(X)) => from(X) activate(n!6220!6220take(X, Y)) => take(X, Y) activate(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]): from(X) >? cons(X, n!6220!6220from(s(X))) head(cons(X, Y)) >? X 2nd(cons(X, Y)) >? head(activate(Y)) take(0, X) >? nil take(s(X), cons(Y, Z)) >? cons(Y, n!6220!6220take(X, activate(Z))) sel(0, cons(X, Y)) >? X sel(s(X), cons(Y, Z)) >? sel(X, activate(Z)) from(X) >? n!6220!6220from(X) take(X, Y) >? n!6220!6220take(X, Y) activate(n!6220!6220from(X)) >? from(X) activate(n!6220!6220take(X, Y)) >? take(X, Y) activate(X) >? X about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[nil]] = _|_ We choose Lex = {sel} and Mul = {0, 2nd, activate, cons, from, head, n!6220!6220from, n!6220!6220take, s, take}, and the following precedence: 2nd > head > sel > activate = from = take > n!6220!6220from > s > n!6220!6220take > 0 > cons Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: from(X) >= cons(X, n!6220!6220from(s(X))) head(cons(X, Y)) >= X 2nd(cons(X, Y)) >= head(activate(Y)) take(0, X) > _|_ take(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220take(X, activate(Z))) sel(0, cons(X, Y)) >= X sel(s(X), cons(Y, Z)) > sel(X, activate(Z)) from(X) >= n!6220!6220from(X) take(X, Y) > n!6220!6220take(X, Y) activate(n!6220!6220from(X)) >= from(X) activate(n!6220!6220take(X, Y)) >= take(X, Y) activate(X) > X With these choices, we have: 1] from(X) >= cons(X, n!6220!6220from(s(X))) because [2], by (Star) 2] from*(X) >= cons(X, n!6220!6220from(s(X))) because from > cons, [3] and [5], by (Copy) 3] from*(X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] from*(X) >= n!6220!6220from(s(X)) because from > n!6220!6220from and [6], by (Copy) 6] from*(X) >= s(X) because from > s and [3], by (Copy) 7] head(cons(X, Y)) >= X because [8], by (Star) 8] head*(cons(X, Y)) >= X because [9], by (Select) 9] cons(X, Y) >= X because [10], by (Star) 10] cons*(X, Y) >= X because [4], by (Select) 11] 2nd(cons(X, Y)) >= head(activate(Y)) because [12], by (Star) 12] 2nd*(cons(X, Y)) >= head(activate(Y)) because 2nd > head and [13], by (Copy) 13] 2nd*(cons(X, Y)) >= activate(Y) because 2nd > activate and [14], by (Copy) 14] 2nd*(cons(X, Y)) >= Y because [15], by (Select) 15] cons(X, Y) >= Y because [16], by (Star) 16] cons*(X, Y) >= Y because [17], by (Select) 17] Y >= Y by (Meta)
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