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HRS union beta 16688 pair #381734581
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
n042.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.103598117828 seconds
cpu usage
0.093076413
max memory
4456448.0
stage attributes
key
value
output-size
4799
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: !plus : [nat * nat] --> nat cons : [nat * list] --> list map : [nat -> nat * list] --> list nil : [] --> list ps : [list] --> list Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) ps(nil) => nil ps(cons(x, y)) => cons(x, ps(map(/\z.!plus(x, z), y))) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] map#(F, cons(X, Y)) =#> map#(F, Y) 1] ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, x), Y)) 2] ps#(cons(X, Y)) =#> map#(/\x.!plus(X, x), Y) Rules R_0: map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) ps(nil) => nil ps(cons(X, Y)) => cons(X, ps(map(/\x.!plus(X, x), Y))) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0 * 1 : 1, 2 * 2 : 0 This graph has the following strongly connected components: P_1: map#(F, cons(X, Y)) =#> map#(F, Y) P_2: ps#(cons(X, Y)) =#> ps#(map(/\x.!plus(X, x), Y)) 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). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). The formative rules of (P_2, R_0) are R_1 ::= map(F, cons(X, Y)) => cons(F X, map(F, Y)) ps(cons(X, Y)) => cons(X, ps(map(/\x.!plus(X, x), Y))) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_2, R_0, minimal, formative) by (P_2, R_1, minimal, formative). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_1, minimal, formative). 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: ps#(cons(X, Y)) >? ps#(map(/\x.!plus(X, x), Y)) map(F, cons(X, Y)) >= cons(F X, map(F, Y)) ps(cons(X, Y)) >= cons(X, ps(map(/\x.!plus(X, x), Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !plus = \y0y1.0 cons = \y0y1.1 + y1 map = \G0y1.y1 ps = \y0.y0 ps# = \y0.2y0 Using this interpretation, the requirements translate to: [[ps#(cons(_x0, _x1))]] = 2 + 2x1 > 2x1 = [[ps#(map(/\x.!plus(_x0, x), _x1))]] [[map(_F0, cons(_x1, _x2))]] = 1 + x2 >= 1 + x2 = [[cons(_F0 _x1, map(_F0, _x2))]]
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