Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Stand 20472 pair #381714933
details
property
value
status
complete
benchmark
2.21.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n014.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0807249546051 seconds
cpu usage
0.077129079
max memory
3416064.0
stage attributes
key
value
output-size
4047
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. !plus : [o * o] --> o 0 : [] --> o bin : [o * o] --> o s : [o] --> o bin(X, 0) => s(0) bin(0, s(X)) => 0 bin(s(X), s(Y)) => !plus(bin(X, s(Y)), bin(X, 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]): bin(X, 0) >? s(0) bin(0, s(X)) >? 0 bin(s(X), s(Y)) >? !plus(bin(X, s(Y)), bin(X, Y)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {!plus, 0, bin, s}, and the following precedence: bin > !plus > 0 > s With these choices, we have: 1] bin(X, 0) > s(0) because [2], by definition 2] bin*(X, 0) >= s(0) because bin > s and [3], by (Copy) 3] bin*(X, 0) >= 0 because bin > 0, by (Copy) 4] bin(0, s(X)) >= 0 because [5], by (Star) 5] bin*(0, s(X)) >= 0 because bin > 0, by (Copy) 6] bin(s(X), s(Y)) >= !plus(bin(X, s(Y)), bin(X, Y)) because [7], by (Star) 7] bin*(s(X), s(Y)) >= !plus(bin(X, s(Y)), bin(X, Y)) because bin > !plus, [8] and [14], by (Copy) 8] bin*(s(X), s(Y)) >= bin(X, s(Y)) because bin in Mul, [9] and [12], by (Stat) 9] s(X) > X because [10], by definition 10] s*(X) >= X because [11], by (Select) 11] X >= X by (Meta) 12] s(Y) >= s(Y) because s in Mul and [13], by (Fun) 13] Y >= Y by (Meta) 14] bin*(s(X), s(Y)) >= bin(X, Y) because bin in Mul, [15] and [16], by (Stat) 15] s(X) >= X because [10], by (Star) 16] s(Y) > Y because [17], by definition 17] s*(Y) >= Y because [13], by (Select) We can thus remove the following rules: bin(X, 0) => s(0) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): bin(0, s(X)) >? 0 bin(s(X), s(Y)) >? !plus(bin(X, s(Y)), bin(X, Y)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {!plus, 0, bin, s}, and the following precedence: bin > !plus > 0 > s With these choices, we have: 1] bin(0, s(X)) >= 0 because [2], by (Star) 2] bin*(0, s(X)) >= 0 because bin > 0, by (Copy) 3] bin(s(X), s(Y)) > !plus(bin(X, s(Y)), bin(X, Y)) because [4], by definition 4] bin*(s(X), s(Y)) >= !plus(bin(X, s(Y)), bin(X, Y)) because bin > !plus, [5] and [11], by (Copy) 5] bin*(s(X), s(Y)) >= bin(X, s(Y)) because bin in Mul, [6] and [9], by (Stat) 6] s(X) > X because [7], by definition 7] s*(X) >= X because [8], by (Select) 8] X >= X by (Meta) 9] s(Y) >= s(Y) because s in Mul and [10], by (Fun) 10] Y >= Y by (Meta) 11] bin*(s(X), s(Y)) >= bin(X, Y) because bin in Mul, [6] and [12], by (Stat) 12] s(Y) >= Y because [13], by (Star) 13] s*(Y) >= Y because [10], by (Select) We can thus remove the following rules: bin(s(X), s(Y)) => !plus(bin(X, s(Y)), bin(X, 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]): bin(0, s(X)) >? 0
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Stand 20472