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Higher Order Rewriting Union Beta pair #487094005
details
property
value
status
complete
benchmark
Applicative_05__TreeMap.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n142.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.525198 seconds
cpu usage
0.525621
user time
0.478053
system time
0.047568
max virtual memory
113188.0
max residence set size
16948.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: cons : [c * b] --> b map : [c -> c * b] --> b nil : [] --> b node : [a * b] --> c treemap : [a -> a] --> c -> c Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) treemap(f) node(x, y) => node(f x, map(treemap(f), 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]): map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) treemap(F) node(X, Y) >? node(F X, map(treemap(F), Y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[nil]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, cons, map, node, treemap}, and the following precedence: treemap > @_{o -> o} = map > cons > node Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: map(F, _|_) > _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) @_{o -> o}(treemap(F), node(X, Y)) >= node(@_{o -> o}(F, X), map(treemap(F), Y)) With these choices, we have: 1] map(F, _|_) > _|_ because [2], by definition 2] map*(F, _|_) >= _|_ by (Bot) 3] map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because [4], by (Star) 4] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [5] and [10], by (Copy) 5] map*(F, cons(X, Y)) >= @_{o -> o}(F, X) because map = @_{o -> o}, map in Mul, [6] and [7], by (Stat) 6] F >= F by (Meta) 7] cons(X, Y) > X because [8], by definition 8] cons*(X, Y) >= X because [9], by (Select) 9] X >= X by (Meta) 10] map*(F, cons(X, Y)) >= map(F, Y) because map in Mul, [6] and [11], by (Stat) 11] cons(X, Y) > Y because [12], by definition 12] cons*(X, Y) >= Y because [13], by (Select) 13] Y >= Y by (Meta) 14] @_{o -> o}(treemap(F), node(X, Y)) >= node(@_{o -> o}(F, X), map(treemap(F), Y)) because [15], by (Star) 15] @_{o -> o}*(treemap(F), node(X, Y)) >= node(@_{o -> o}(F, X), map(treemap(F), Y)) because [16], by (Select) 16] treemap(F) @_{o -> o}*(treemap(F), node(X, Y)) >= node(@_{o -> o}(F, X), map(treemap(F), Y)) because [17] 17] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= node(@_{o -> o}(F, X), map(treemap(F), Y)) because treemap > node, [18] and [29], by (Copy) 18] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= @_{o -> o}(F, X) because treemap > @_{o -> o}, [19] and [21], by (Copy) 19] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= F because [20], by (Select) 20] F >= F by (Meta) 21] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= X because [22], by (Select) 22] @_{o -> o}*(treemap(F), node(X, Y)) >= X because [23], by (Select) 23] treemap(F) @_{o -> o}*(treemap(F), node(X, Y)) >= X because [24] 24] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= X because [25], by (Select) 25] @_{o -> o}*(treemap(F), node(X, Y)) >= X because [26], by (Select) 26] node(X, Y) >= X because [27], by (Star) 27] node*(X, Y) >= X because [28], by (Select) 28] X >= X by (Meta) 29] treemap*(F, @_{o -> o}*(treemap(F), node(X, Y))) >= map(treemap(F), Y) because [30], by (Select) 30] @_{o -> o}*(treemap(F), node(X, Y)) >= map(treemap(F), Y) because @_{o -> o} = map, @_{o -> o} in Mul, [31] and [33], by (Stat) 31] treemap(F) >= treemap(F) because treemap in Mul and [32], by (Fun) 32] F >= F by (Meta) 33] node(X, Y) > Y because [34], by definition 34] node*(X, Y) >= Y because [35], by (Select) 35] Y >= Y by (Meta) We can thus remove the following rules: map(F, nil) => nil We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): map(F, cons(X, Y)) >? cons(F X, map(F, Y)) treemap(F) node(X, Y) >? node(F X, map(treemap(F), Y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {@_{o -> o}, cons, map, node, treemap}, and the following precedence: @_{o -> o} = map > cons > node > treemap With these choices, we have: 1] map(F, cons(X, Y)) > cons(@_{o -> o}(F, X), map(F, Y)) because [2], by definition 2] map*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) because map > cons, [3] and [8], by (Copy)
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