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TRS Stand 20472 pair #381711081
details
property
value
status
complete
benchmark
prov.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n050.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0552170276642 seconds
cpu usage
0.051546716
max memory
2510848.0
stage attributes
key
value
output-size
4112
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. ackin : [o * o] --> o ackout : [o] --> o s : [o] --> o u21 : [o * o] --> o u22 : [o] --> o ackin(s(X), s(Y)) => u21(ackin(s(X), Y), X) u21(ackout(X), Y) => u22(ackin(Y, X)) As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: ackin : [i * i] --> ca ackout : [i] --> ca s : [i] --> i u21 : [ca * i] --> ca u22 : [ca] --> ca We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): ackin(s(X), s(Y)) >? u21(ackin(s(X), Y), X) u21(ackout(X), Y) >? u22(ackin(Y, X)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[u21(x_1, x_2)]] = u21(x_2, x_1) [[u22(x_1)]] = x_1 We choose Lex = {ackin, u21} and Mul = {ackout, s}, and the following precedence: ackout > s > ackin = u21 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: ackin(s(X), s(Y)) > u21(ackin(s(X), Y), X) u21(ackout(X), Y) >= ackin(Y, X) With these choices, we have: 1] ackin(s(X), s(Y)) > u21(ackin(s(X), Y), X) because [2], by definition 2] ackin*(s(X), s(Y)) >= u21(ackin(s(X), Y), X) because ackin = u21, [3], [6] and [15], by (Stat) 3] s(X) > X because [4], by definition 4] s*(X) >= X because [5], by (Select) 5] X >= X by (Meta) 6] ackin*(s(X), s(Y)) >= ackin(s(X), Y) because [7], [9], [12] and [13], by (Stat) 7] s(X) >= s(X) because s in Mul and [8], by (Fun) 8] X >= X by (Meta) 9] s(Y) > Y because [10], by definition 10] s*(Y) >= Y because [11], by (Select) 11] Y >= Y by (Meta) 12] ackin*(s(X), s(Y)) >= s(X) because [7], by (Select) 13] ackin*(s(X), s(Y)) >= Y because [14], by (Select) 14] s(Y) >= Y because [10], by (Star) 15] ackin*(s(X), s(Y)) >= X because [16], by (Select) 16] s(X) >= X because [4], by (Star) 17] u21(ackout(X), Y) >= ackin(Y, X) because u21 = ackin, [18] and [20], by (Fun) 18] ackout(X) >= X because [19], by (Star) 19] ackout*(X) >= X because [8], by (Select) 20] Y >= Y by (Meta) We can thus remove the following rules: ackin(s(X), s(Y)) => u21(ackin(s(X), Y), 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]): u21(ackout(X), Y) >? u22(ackin(Y, X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: ackin = \y0y1.y0 + y1 ackout = \y0.3 + 3y0 u21 = \y0y1.3 + 3y0 + 3y1 u22 = \y0.y0 Using this interpretation, the requirements translate to: [[u21(ackout(_x0), _x1)]] = 12 + 3x1 + 9x0 > x0 + x1 = [[u22(ackin(_x1, _x0))]] We can thus remove the following rules:
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