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TRS Stand 20472 pair #381713978
details
property
value
status
complete
benchmark
4.47.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n026.star.cs.uiowa.edu
space
SK90
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0966210365295 seconds
cpu usage
0.094199132
max memory
5910528.0
stage attributes
key
value
output-size
4188
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. !dot : [o * o] --> o a : [] --> o b : [] --> o b!450 : [] --> o c : [] --> o d : [] --> o d!450 : [] --> o e : [] --> o f : [o * o] --> o g : [o * o] --> o h : [o * o] --> o i : [o * o * o] --> o if : [o * o * o] --> o f(g(i(a, b, b!450), c), d) => if(e, f(!dot(b, c), d!450), f(!dot(b!450, c), d!450)) f(g(h(a, b), c), d) => if(e, f(!dot(b, g(h(a, b), c)), d), f(c, d!450)) As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: !dot : [k * ya] --> ya a : [] --> g b : [] --> k b!450 : [] --> k c : [] --> ya d : [] --> aa d!450 : [] --> aa e : [] --> u f : [ya * aa] --> eb g : [va * ya] --> ya h : [g * k] --> va i : [g * k * k] --> va if : [u * eb * eb] --> eb We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): f(g(i(a, b, b!450), c), d) >? if(e, f(!dot(b, c), d!450), f(!dot(b!450, c), d!450)) f(g(h(a, b), c), d) >? if(e, f(!dot(b, g(h(a, b), c)), d), f(c, d!450)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !dot = \y0y1.y0 + y1 a = 3 b = 0 b!450 = 0 c = 0 d = 3 d!450 = 0 e = 0 f = \y0y1.y0 + 3y1 g = \y0y1.3 + 3y0 + 3y1 h = \y0y1.y0 + y1 i = \y0y1y2.3 + 3y0 + 3y1 + 3y2 if = \y0y1y2.y0 + y1 + y2 Using this interpretation, the requirements translate to: [[f(g(i(a, b, b!450), c), d)]] = 48 > 0 = [[if(e, f(!dot(b, c), d!450), f(!dot(b!450, c), d!450))]] [[f(g(h(a, b), c), d)]] = 21 >= 21 = [[if(e, f(!dot(b, g(h(a, b), c)), d), f(c, d!450))]] We can thus remove the following rules: f(g(i(a, b, b!450), c), d) => if(e, f(!dot(b, c), d!450), f(!dot(b!450, c), d!450)) 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] f#(g(h(a, b), c), d) =#> f#(!dot(b, g(h(a, b), c)), d) 1] f#(g(h(a, b), c), d) =#> f#(c, d!450) Rules R_0: f(g(h(a, b), c), d) => if(e, f(!dot(b, g(h(a, b), c)), d), f(c, d!450)) 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 : * 1 :
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