Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Stand 20472 pair #381714385
details
property
value
status
complete
benchmark
4.09.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n030.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.140421867371 seconds
cpu usage
0.137847169
max memory
6119424.0
stage attributes
key
value
output-size
8603
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. !plus : [o * o] --> o !times : [o * o] --> o 0 : [] --> o i : [o] --> o !plus(X, 0) => X !plus(X, i(X)) => 0 !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) !times(X, !plus(Y, Z)) => !plus(!times(X, Y), !times(X, Z)) !times(!plus(X, Y), Z) => !plus(!times(X, Z), !times(Y, Z)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): !plus(X, 0) >? X !plus(X, i(X)) >? 0 !plus(!plus(X, Y), Z) >? !plus(X, !plus(Y, Z)) !times(X, !plus(Y, Z)) >? !plus(!times(X, Y), !times(X, Z)) !times(!plus(X, Y), Z) >? !plus(!times(X, Z), !times(Y, Z)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ We choose Lex = {!plus} and Mul = {!times, i}, and the following precedence: i > !times > !plus Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: !plus(X, _|_) >= X !plus(X, i(X)) > _|_ !plus(!plus(X, Y), Z) > !plus(X, !plus(Y, Z)) !times(X, !plus(Y, Z)) >= !plus(!times(X, Y), !times(X, Z)) !times(!plus(X, Y), Z) >= !plus(!times(X, Z), !times(Y, Z)) With these choices, we have: 1] !plus(X, _|_) >= X because [2], by (Star) 2] !plus*(X, _|_) >= X because [3], by (Select) 3] X >= X by (Meta) 4] !plus(X, i(X)) > _|_ because [5], by definition 5] !plus*(X, i(X)) >= _|_ by (Bot) 6] !plus(!plus(X, Y), Z) > !plus(X, !plus(Y, Z)) because [7], by definition 7] !plus*(!plus(X, Y), Z) >= !plus(X, !plus(Y, Z)) because [8], [10] and [12], by (Stat) 8] !plus(X, Y) > X because [9], by definition 9] !plus*(X, Y) >= X because [3], by (Select) 10] !plus*(!plus(X, Y), Z) >= X because [11], by (Select) 11] !plus(X, Y) >= X because [9], by (Star) 12] !plus*(!plus(X, Y), Z) >= !plus(Y, Z) because [13], [16] and [18], by (Stat) 13] !plus(X, Y) > Y because [14], by definition 14] !plus*(X, Y) >= Y because [15], by (Select) 15] Y >= Y by (Meta) 16] !plus*(!plus(X, Y), Z) >= Y because [17], by (Select) 17] !plus(X, Y) >= Y because [14], by (Star) 18] !plus*(!plus(X, Y), Z) >= Z because [19], by (Select) 19] Z >= Z by (Meta) 20] !times(X, !plus(Y, Z)) >= !plus(!times(X, Y), !times(X, Z)) because [21], by (Star) 21] !times*(X, !plus(Y, Z)) >= !plus(!times(X, Y), !times(X, Z)) because !times > !plus, [22] and [26], by (Copy) 22] !times*(X, !plus(Y, Z)) >= !times(X, Y) because !times in Mul, [23] and [24], by (Stat) 23] X >= X by (Meta) 24] !plus(Y, Z) > Y because [25], by definition 25] !plus*(Y, Z) >= Y because [15], by (Select) 26] !times*(X, !plus(Y, Z)) >= !times(X, Z) because !times in Mul, [23] and [27], by (Stat) 27] !plus(Y, Z) > Z because [28], by definition 28] !plus*(Y, Z) >= Z because [19], by (Select) 29] !times(!plus(X, Y), Z) >= !plus(!times(X, Z), !times(Y, Z)) because [30], by (Star) 30] !times*(!plus(X, Y), Z) >= !plus(!times(X, Z), !times(Y, Z)) because !times > !plus, [31] and [33], by (Copy) 31] !times*(!plus(X, Y), Z) >= !times(X, Z) because !times in Mul, [8] and [32], by (Stat) 32] Z >= Z by (Meta) 33] !times*(!plus(X, Y), Z) >= !times(Y, Z) because !times in Mul, [13] and [32], by (Stat) We can thus remove the following rules: !plus(X, i(X)) => 0 !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): !plus(X, 0) >? X
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Stand 20472