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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270313
details
property
value
status
complete
benchmark
prefixsum.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n133.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.202283 seconds
cpu usage
0.198882
user time
0.156373
system time
0.042509
max virtual memory
113176.0
max residence set size
5700.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.19 YES 0.00/0.20 We consider the system theBenchmark. 0.00/0.20 0.00/0.20 Alphabet: 0.00/0.20 0.00/0.20 !plus : [nat * nat] --> nat 0.00/0.20 cons : [nat * list] --> list 0.00/0.20 map : [nat -> nat * list] --> list 0.00/0.20 nil : [] --> list 0.00/0.20 ps : [list] --> list 0.00/0.20 0.00/0.20 Rules: 0.00/0.20 0.00/0.20 map(f, nil) => nil 0.00/0.20 map(f, cons(x, y)) => cons(f x, map(f, y)) 0.00/0.20 ps(nil) => nil 0.00/0.20 ps(cons(x, y)) => cons(x, ps(map(/\z.!plus(x, z), y))) 0.00/0.20 0.00/0.20 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.20 0.00/0.20 We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]). 0.00/0.20 0.00/0.20 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 0.00/0.20 0.00/0.20 Dependency Pairs P_0: 0.00/0.20 0.00/0.20 0] map#(F, cons(X, Y)) =#> map#(F, Y) 0.00/0.20 1] ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, x), Y)) 0.00/0.20 2] ps#(cons(X, Y)) =#> map#(/\x.!plus(X, x), Y) 0.00/0.20 0.00/0.20 Rules R_0: 0.00/0.20 0.00/0.20 map(F, nil) => nil 0.00/0.20 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 0.00/0.20 ps(nil) => nil 0.00/0.20 ps(cons(X, Y)) => cons(X, ps(map(/\x.!plus(X, x), Y))) 0.00/0.20 0.00/0.20 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 0.00/0.20 0.00/0.20 We consider the dependency pair problem (P_0, R_0, static, formative). 0.00/0.20 0.00/0.20 We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: 0.00/0.20 0.00/0.20 * 0 : 0 0.00/0.20 * 1 : 1, 2 0.00/0.20 * 2 : 0 0.00/0.20 0.00/0.20 This graph has the following strongly connected components: 0.00/0.20 0.00/0.20 P_1: 0.00/0.20 0.00/0.20 map#(F, cons(X, Y)) =#> map#(F, Y) 0.00/0.20 0.00/0.20 P_2: 0.00/0.20 0.00/0.20 ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, x), Y)) 0.00/0.20 0.00/0.20 By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f) and (P_2, R_0, m, f). 0.00/0.20 0.00/0.20 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_0, static, formative) is finite. 0.00/0.20 0.00/0.20 We consider the dependency pair problem (P_2, R_0, static, formative). 0.00/0.20 0.00/0.20 The formative rules of (P_2, R_0) are R_1 ::= 0.00/0.20 0.00/0.20 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 0.00/0.20 ps(cons(X, Y)) => cons(X, ps(map(/\x.!plus(X, x), Y))) 0.00/0.20 0.00/0.20 By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, static, formative) by (P_2, R_1, static, formative). 0.00/0.20 0.00/0.20 Thus, the original system is terminating if each of (P_1, R_0, static, formative) and (P_2, R_1, static, formative) is finite. 0.00/0.20 0.00/0.20 We consider the dependency pair problem (P_2, R_1, static, formative). 0.00/0.20 0.00/0.20 We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: 0.00/0.20 0.00/0.20 ps#(cons(X, Y)) >? ps#(map(/\x.!plus(X, x), Y)) 0.00/0.20 map(F, cons(X, Y)) >= cons(F X, map(F, Y)) 0.00/0.20 ps(cons(X, Y)) >= cons(X, ps(map(/\x.!plus(X, x), Y))) 0.00/0.20 0.00/0.20 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.20 0.00/0.20 The following interpretation satisfies the requirements: 0.00/0.20 0.00/0.20 !plus = \y0y1.0 0.00/0.20 cons = \y0y1.1 + y1 0.00/0.20 map = \G0y1.y1 0.00/0.20 ps = \y0.y0 0.00/0.20 ps# = \y0.2y0 0.00/0.20 0.00/0.20 Using this interpretation, the requirements translate to: 0.00/0.20 0.00/0.20 [[ps#(cons(_x0, _x1))]] = 2 + 2x1 > 2x1 = [[ps#(map(/\x.!plus(_x0, x), _x1))]] 0.00/0.20 [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]] 0.00/0.20 [[ps(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[cons(_x0, ps(map(/\x.!plus(_x0, x), _x1)))]] 0.00/0.20 0.00/0.20 By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_2, R_1) by ({}, R_1). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. 0.00/0.20 0.00/0.20 Thus, the original system is terminating if (P_1, R_0, static, formative) is finite. 0.00/0.20
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