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TRS Stand 20472 pair #381714038
details
property
value
status
complete
benchmark
2.07.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n071.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.099326133728 seconds
cpu usage
0.096692958
max memory
4534272.0
stage attributes
key
value
output-size
6755
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 1 : [] --> o 2 : [] --> o f : [o * o] --> o g : [o * o] --> o i : [o] --> o f(0, X) => X f(X, 0) => X f(i(X), Y) => i(X) f(f(X, Y), Z) => f(X, f(Y, Z)) f(g(X, Y), Z) => g(f(X, Z), f(Y, Z)) f(1, g(X, Y)) => X f(2, g(X, Y)) => 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]): f(0, X) >? X f(X, 0) >? X f(i(X), Y) >? i(X) f(f(X, Y), Z) >? f(X, f(Y, Z)) f(g(X, Y), Z) >? g(f(X, Z), f(Y, Z)) f(1, g(X, Y)) >? X f(2, g(X, Y)) >? Y about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {f} and Mul = {0, 1, 2, g, i}, and the following precedence: i > 1 > 0 > f > g > 2 With these choices, we have: 1] f(0, X) >= X because [2], by (Star) 2] f*(0, X) >= X because [3], by (Select) 3] X >= X by (Meta) 4] f(X, 0) >= X because [5], by (Star) 5] f*(X, 0) >= X because [6], by (Select) 6] X >= X by (Meta) 7] f(i(X), Y) > i(X) because [8], by definition 8] f*(i(X), Y) >= i(X) because [9], by (Select) 9] i(X) >= i(X) because i in Mul and [10], by (Fun) 10] X >= X by (Meta) 11] f(f(X, Y), Z) >= f(X, f(Y, Z)) because [12], by (Star) 12] f*(f(X, Y), Z) >= f(X, f(Y, Z)) because [13], [15] and [17], by (Stat) 13] f(X, Y) > X because [14], by definition 14] f*(X, Y) >= X because [10], by (Select) 15] f*(f(X, Y), Z) >= X because [16], by (Select) 16] f(X, Y) >= X because [14], by (Star) 17] f*(f(X, Y), Z) >= f(Y, Z) because [18], [20] and [22], by (Stat) 18] f(X, Y) > Y because [19], by definition 19] f*(X, Y) >= Y because [3], by (Select) 20] f*(f(X, Y), Z) >= Y because [21], by (Select) 21] f(X, Y) >= Y because [19], by (Star) 22] f*(f(X, Y), Z) >= Z because [23], by (Select) 23] Z >= Z by (Meta) 24] f(g(X, Y), Z) >= g(f(X, Z), f(Y, Z)) because [25], by (Star) 25] f*(g(X, Y), Z) >= g(f(X, Z), f(Y, Z)) because f > g, [26] and [32], by (Copy) 26] f*(g(X, Y), Z) >= f(X, Z) because [27], [29] and [31], by (Stat) 27] g(X, Y) > X because [28], by definition 28] g*(X, Y) >= X because [10], by (Select) 29] f*(g(X, Y), Z) >= X because [30], by (Select) 30] g(X, Y) >= X because [28], by (Star) 31] f*(g(X, Y), Z) >= Z because [23], by (Select) 32] f*(g(X, Y), Z) >= f(Y, Z) because [33], [35] and [31], by (Stat) 33] g(X, Y) > Y because [34], by definition 34] g*(X, Y) >= Y because [3], by (Select) 35] f*(g(X, Y), Z) >= Y because [36], by (Select) 36] g(X, Y) >= Y because [34], by (Star) 37] f(1, g(X, Y)) > X because [38], by definition 38] f*(1, g(X, Y)) >= X because [30], by (Select) 39] f(2, g(X, Y)) >= Y because [40], by (Star) 40] f*(2, g(X, Y)) >= Y because [36], by (Select) We can thus remove the following rules: f(i(X), Y) => i(X) f(1, g(X, Y)) => X We use rule removal, following [Kop12, Theorem 2.23].
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