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TRS Stand 20472 pair #381715175
details
property
value
status
complete
benchmark
Ex3_12_Luc96a_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n051.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0704910755157 seconds
cpu usage
0.06619227
max memory
2748416.0
stage attributes
key
value
output-size
5265
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 activate : [o] --> o cons : [o * o] --> o from : [o] --> o n!6220!6220from : [o] --> o s : [o] --> o sel : [o * o] --> o from(X) => cons(X, n!6220!6220from(s(X))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) from(X) => n!6220!6220from(X) activate(n!6220!6220from(X)) => from(X) 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))) sel(0, cons(X, Y)) >? X sel(s(X), cons(Y, Z)) >? sel(X, activate(Z)) from(X) >? n!6220!6220from(X) activate(n!6220!6220from(X)) >? from(X) activate(X) >? X about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {sel} and Mul = {0, activate, cons, from, n!6220!6220from, s}, and the following precedence: 0 > sel > activate > from > cons > n!6220!6220from > s With these choices, we have: 1] from(X) > cons(X, n!6220!6220from(s(X))) because [2], by definition 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] sel(0, cons(X, Y)) >= X because [8], by (Star) 8] sel*(0, 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] sel(s(X), cons(Y, Z)) >= sel(X, activate(Z)) because [12], by (Star) 12] sel*(s(X), cons(Y, Z)) >= sel(X, activate(Z)) because [13], [15] and [17], by (Stat) 13] s(X) > X because [14], by definition 14] s*(X) >= X because [4], by (Select) 15] sel*(s(X), cons(Y, Z)) >= X because [16], by (Select) 16] s(X) >= X because [14], by (Star) 17] sel*(s(X), cons(Y, Z)) >= activate(Z) because sel > activate and [18], by (Copy) 18] sel*(s(X), cons(Y, Z)) >= Z because [19], by (Select) 19] cons(Y, Z) >= Z because [20], by (Star) 20] cons*(Y, Z) >= Z because [21], by (Select) 21] Z >= Z by (Meta) 22] from(X) >= n!6220!6220from(X) because [23], by (Star) 23] from*(X) >= n!6220!6220from(X) because from > n!6220!6220from and [3], by (Copy) 24] activate(n!6220!6220from(X)) >= from(X) because [25], by (Star) 25] activate*(n!6220!6220from(X)) >= from(X) because activate > from and [26], by (Copy) 26] activate*(n!6220!6220from(X)) >= X because [27], by (Select) 27] n!6220!6220from(X) >= X because [28], by (Star) 28] n!6220!6220from*(X) >= X because [4], by (Select) 29] activate(X) > X because [30], by definition 30] activate*(X) >= X because [4], by (Select) We can thus remove the following rules: from(X) => cons(X, n!6220!6220from(s(X))) 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]): sel(0, cons(X, Y)) >? X sel(s(X), cons(Y, Z)) >? sel(X, activate(Z)) from(X) >? n!6220!6220from(X) activate(n!6220!6220from(X)) >? from(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3
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