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HRS union beta 16688 pair #381734197
details
property
value
status
complete
benchmark
02Ackermann.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n004.star.cs.uiowa.edu
space
Blanqui_15
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.10791516304 seconds
cpu usage
0.104503936
max memory
4259840.0
stage attributes
key
value
output-size
6369
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: ack : [N * N] --> N s : [N] --> N z : [] --> N Rules: ack(z, x) => s(x) ack(s(x), z) => ack(x, s(z)) ack(s(x), s(y)) => ack(x, ack(s(x), y)) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 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]): ack(z, X) >? s(X) ack(s(X), z) >? ack(X, s(z)) ack(s(X), s(Y)) >? ack(X, ack(s(X), Y)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[z]] = _|_ We choose Lex = {ack} and Mul = {s}, and the following precedence: ack > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: ack(_|_, X) > s(X) ack(s(X), _|_) >= ack(X, s(_|_)) ack(s(X), s(Y)) >= ack(X, ack(s(X), Y)) With these choices, we have: 1] ack(_|_, X) > s(X) because [2], by definition 2] ack*(_|_, X) >= s(X) because ack > s and [3], by (Copy) 3] ack*(_|_, X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] ack(s(X), _|_) >= ack(X, s(_|_)) because [6], by (Star) 6] ack*(s(X), _|_) >= ack(X, s(_|_)) because [7], [10] and [12], by (Stat) 7] s(X) > X because [8], by definition 8] s*(X) >= X because [9], by (Select) 9] X >= X by (Meta) 10] ack*(s(X), _|_) >= X because [11], by (Select) 11] s(X) >= X because [8], by (Star) 12] ack*(s(X), _|_) >= s(_|_) because ack > s and [13], by (Copy) 13] ack*(s(X), _|_) >= _|_ by (Bot) 14] ack(s(X), s(Y)) >= ack(X, ack(s(X), Y)) because [15], by (Star) 15] ack*(s(X), s(Y)) >= ack(X, ack(s(X), Y)) because [16], [19] and [21], by (Stat) 16] s(X) > X because [17], by definition 17] s*(X) >= X because [18], by (Select) 18] X >= X by (Meta) 19] ack*(s(X), s(Y)) >= X because [20], by (Select) 20] s(X) >= X because [17], by (Star) 21] ack*(s(X), s(Y)) >= ack(s(X), Y) because [22], [24], [27] and [28], by (Stat) 22] s(X) >= s(X) because s in Mul and [23], by (Fun) 23] X >= X by (Meta) 24] s(Y) > Y because [25], by definition 25] s*(Y) >= Y because [26], by (Select) 26] Y >= Y by (Meta) 27] ack*(s(X), s(Y)) >= s(X) because ack > s and [19], by (Copy) 28] ack*(s(X), s(Y)) >= Y because [29], by (Select) 29] s(Y) >= Y because [25], by (Star) We can thus remove the following rules: ack(z, X) => s(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]): ack(s(X), z) >? ack(X, s(z)) ack(s(X), s(Y)) >? ack(X, ack(s(X), Y)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[z]] = _|_
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