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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270491
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
n099.star.cs.uiowa.edu
space
Blanqui_15
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.188873 seconds
cpu usage
0.18609
user time
0.147725
system time
0.038365
max virtual memory
113176.0
max residence set size
5876.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.17 YES 0.00/0.18 We consider the system theBenchmark. 0.00/0.18 0.00/0.18 Alphabet: 0.00/0.18 0.00/0.18 ack : [N * N] --> N 0.00/0.18 s : [N] --> N 0.00/0.18 z : [] --> N 0.00/0.18 0.00/0.18 Rules: 0.00/0.18 0.00/0.18 ack(z, x) => s(x) 0.00/0.18 ack(s(x), z) => ack(x, s(z)) 0.00/0.18 ack(s(x), s(y)) => ack(x, ack(s(x), y)) 0.00/0.18 0.00/0.18 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.18 0.00/0.18 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.18 0.00/0.18 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.18 0.00/0.18 ack(z, X) >? s(X) 0.00/0.18 ack(s(X), z) >? ack(X, s(z)) 0.00/0.18 ack(s(X), s(Y)) >? ack(X, ack(s(X), Y)) 0.00/0.18 0.00/0.18 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.00/0.18 0.00/0.18 Argument functions: 0.00/0.18 0.00/0.18 [[z]] = _|_ 0.00/0.18 0.00/0.18 We choose Lex = {ack} and Mul = {s}, and the following precedence: ack > s 0.00/0.18 0.00/0.18 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.00/0.18 0.00/0.18 ack(_|_, X) > s(X) 0.00/0.18 ack(s(X), _|_) >= ack(X, s(_|_)) 0.00/0.18 ack(s(X), s(Y)) >= ack(X, ack(s(X), Y)) 0.00/0.18 0.00/0.18 With these choices, we have: 0.00/0.18 0.00/0.18 1] ack(_|_, X) > s(X) because [2], by definition 0.00/0.18 2] ack*(_|_, X) >= s(X) because ack > s and [3], by (Copy) 0.00/0.18 3] ack*(_|_, X) >= X because [4], by (Select) 0.00/0.18 4] X >= X by (Meta) 0.00/0.18 0.00/0.18 5] ack(s(X), _|_) >= ack(X, s(_|_)) because [6], by (Star) 0.00/0.18 6] ack*(s(X), _|_) >= ack(X, s(_|_)) because [7], [10] and [12], by (Stat) 0.00/0.18 7] s(X) > X because [8], by definition 0.00/0.18 8] s*(X) >= X because [9], by (Select) 0.00/0.18 9] X >= X by (Meta) 0.00/0.18 10] ack*(s(X), _|_) >= X because [11], by (Select) 0.00/0.18 11] s(X) >= X because [8], by (Star) 0.00/0.18 12] ack*(s(X), _|_) >= s(_|_) because ack > s and [13], by (Copy) 0.00/0.18 13] ack*(s(X), _|_) >= _|_ by (Bot) 0.00/0.18 0.00/0.18 14] ack(s(X), s(Y)) >= ack(X, ack(s(X), Y)) because [15], by (Star) 0.00/0.18 15] ack*(s(X), s(Y)) >= ack(X, ack(s(X), Y)) because [16], [19] and [21], by (Stat) 0.00/0.18 16] s(X) > X because [17], by definition 0.00/0.18 17] s*(X) >= X because [18], by (Select) 0.00/0.18 18] X >= X by (Meta) 0.00/0.18 19] ack*(s(X), s(Y)) >= X because [20], by (Select) 0.00/0.18 20] s(X) >= X because [17], by (Star) 0.00/0.18 21] ack*(s(X), s(Y)) >= ack(s(X), Y) because [22], [24], [27] and [28], by (Stat) 0.00/0.18 22] s(X) >= s(X) because s in Mul and [23], by (Fun) 0.00/0.18 23] X >= X by (Meta) 0.00/0.18 24] s(Y) > Y because [25], by definition 0.00/0.18 25] s*(Y) >= Y because [26], by (Select) 0.00/0.18 26] Y >= Y by (Meta) 0.00/0.18 27] ack*(s(X), s(Y)) >= s(X) because ack > s and [19], by (Copy) 0.00/0.18 28] ack*(s(X), s(Y)) >= Y because [29], by (Select) 0.00/0.18 29] s(Y) >= Y because [25], by (Star) 0.00/0.18 0.00/0.18 We can thus remove the following rules: 0.00/0.18 0.00/0.18 ack(z, X) => s(X) 0.00/0.18 0.00/0.18 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.18 0.00/0.18 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.18 0.00/0.18 ack(s(X), z) >? ack(X, s(z)) 0.00/0.18 ack(s(X), s(Y)) >? ack(X, ack(s(X), Y)) 0.00/0.18 0.00/0.18 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.00/0.18 0.00/0.18 Argument functions: 0.00/0.18 0.00/0.18 [[z]] = _|_ 0.00/0.18 0.00/0.18 We choose Lex = {ack} and Mul = {s}, and the following precedence: ack > s 0.00/0.18 0.00/0.18 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.00/0.18 0.00/0.18 ack(s(X), _|_) > ack(X, s(_|_)) 0.00/0.18 ack(s(X), s(Y)) >= ack(X, ack(s(X), Y)) 0.00/0.18 0.00/0.18 With these choices, we have: 0.00/0.18 0.00/0.18 1] ack(s(X), _|_) > ack(X, s(_|_)) because [2], by definition
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