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TRS Stand 20472 pair #381713587
details
property
value
status
complete
benchmark
#3.16.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n072.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.285757064819 seconds
cpu usage
0.164801535
max memory
5644288.0
stage attributes
key
value
output-size
9204
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 plus : [o * o] --> o s : [o] --> o times : [o * o] --> o times(X, 0) => 0 times(X, s(Y)) => plus(times(X, Y), X) plus(X, 0) => X plus(0, X) => X plus(X, s(Y)) => s(plus(X, Y)) plus(s(X), Y) => s(plus(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]): times(X, 0) >? 0 times(X, s(Y)) >? plus(times(X, Y), X) plus(X, 0) >? X plus(0, X) >? X plus(X, s(Y)) >? s(plus(X, Y)) plus(s(X), Y) >? s(plus(X, Y)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ We choose Lex = {} and Mul = {plus, s, times}, and the following precedence: times > plus > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: times(X, _|_) >= _|_ times(X, s(Y)) >= plus(times(X, Y), X) plus(X, _|_) >= X plus(_|_, X) >= X plus(X, s(Y)) > s(plus(X, Y)) plus(s(X), Y) >= s(plus(X, Y)) With these choices, we have: 1] times(X, _|_) >= _|_ by (Bot) 2] times(X, s(Y)) >= plus(times(X, Y), X) because [3], by (Star) 3] times*(X, s(Y)) >= plus(times(X, Y), X) because times > plus, [4] and [9], by (Copy) 4] times*(X, s(Y)) >= times(X, Y) because times in Mul, [5] and [6], by (Stat) 5] X >= X by (Meta) 6] s(Y) > Y because [7], by definition 7] s*(Y) >= Y because [8], by (Select) 8] Y >= Y by (Meta) 9] times*(X, s(Y)) >= X because [5], by (Select) 10] plus(X, _|_) >= X because [11], by (Star) 11] plus*(X, _|_) >= X because [5], by (Select) 12] plus(_|_, X) >= X because [13], by (Star) 13] plus*(_|_, X) >= X because [5], by (Select) 14] plus(X, s(Y)) > s(plus(X, Y)) because [15], by definition 15] plus*(X, s(Y)) >= s(plus(X, Y)) because plus > s and [16], by (Copy) 16] plus*(X, s(Y)) >= plus(X, Y) because plus in Mul, [5] and [6], by (Stat) 17] plus(s(X), Y) >= s(plus(X, Y)) because [18], by (Star) 18] plus*(s(X), Y) >= s(plus(X, Y)) because plus > s and [19], by (Copy) 19] plus*(s(X), Y) >= plus(X, Y) because plus in Mul, [20] and [22], by (Stat) 20] s(X) > X because [21], by definition 21] s*(X) >= X because [5], by (Select) 22] Y >= Y by (Meta) We can thus remove the following rules: plus(X, s(Y)) => s(plus(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]): times(X, 0) >? 0 times(X, s(Y)) >? plus(times(X, Y), X) plus(X, 0) >? X plus(0, X) >? X plus(s(X), Y) >? s(plus(X, Y)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5].
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