Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Stand 20472 pair #381715560
details
property
value
status
complete
benchmark
Ex4_4_Luc96b_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n102.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0918231010437 seconds
cpu usage
0.074367439
max memory
3379200.0
stage attributes
key
value
output-size
4846
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. a!6220!6220f : [o * o] --> o f : [o * o] --> o g : [o] --> o mark : [o] --> o a!6220!6220f(g(X), Y) => a!6220!6220f(mark(X), f(g(X), Y)) mark(f(X, Y)) => a!6220!6220f(mark(X), Y) mark(g(X)) => g(mark(X)) a!6220!6220f(X, Y) => f(X, Y) 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] a!6220!6220f#(g(X), Y) =#> a!6220!6220f#(mark(X), f(g(X), Y)) 1] a!6220!6220f#(g(X), Y) =#> mark#(X) 2] mark#(f(X, Y)) =#> a!6220!6220f#(mark(X), Y) 3] mark#(f(X, Y)) =#> mark#(X) 4] mark#(g(X)) =#> mark#(X) Rules R_0: a!6220!6220f(g(X), Y) => a!6220!6220f(mark(X), f(g(X), Y)) mark(f(X, Y)) => a!6220!6220f(mark(X), Y) mark(g(X)) => g(mark(X)) a!6220!6220f(X, Y) => f(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 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: a!6220!6220f#(g(X), Y) >? a!6220!6220f#(mark(X), f(g(X), Y)) a!6220!6220f#(g(X), Y) >? mark#(X) mark#(f(X, Y)) >? a!6220!6220f#(mark(X), Y) mark#(f(X, Y)) >? mark#(X) mark#(g(X)) >? mark#(X) a!6220!6220f(g(X), Y) >= a!6220!6220f(mark(X), f(g(X), Y)) mark(f(X, Y)) >= a!6220!6220f(mark(X), Y) mark(g(X)) >= g(mark(X)) a!6220!6220f(X, Y) >= f(X, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a!6220!6220f = \y0y1.2y0 a!6220!6220f# = \y0y1.y0 f = \y0y1.2y0 g = \y0.1 + 2y0 mark = \y0.2y0 mark# = \y0.y0 Using this interpretation, the requirements translate to: [[a!6220!6220f#(g(_x0), _x1)]] = 1 + 2x0 > 2x0 = [[a!6220!6220f#(mark(_x0), f(g(_x0), _x1))]] [[a!6220!6220f#(g(_x0), _x1)]] = 1 + 2x0 > x0 = [[mark#(_x0)]] [[mark#(f(_x0, _x1))]] = 2x0 >= 2x0 = [[a!6220!6220f#(mark(_x0), _x1)]] [[mark#(f(_x0, _x1))]] = 2x0 >= x0 = [[mark#(_x0)]] [[mark#(g(_x0))]] = 1 + 2x0 > x0 = [[mark#(_x0)]] [[a!6220!6220f(g(_x0), _x1)]] = 2 + 4x0 >= 4x0 = [[a!6220!6220f(mark(_x0), f(g(_x0), _x1))]] [[mark(f(_x0, _x1))]] = 4x0 >= 4x0 = [[a!6220!6220f(mark(_x0), _x1)]] [[mark(g(_x0))]] = 2 + 4x0 >= 1 + 4x0 = [[g(mark(_x0))]] [[a!6220!6220f(_x0, _x1)]] = 2x0 >= 2x0 = [[f(_x0, _x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_1, R_0, minimal, formative), where P_1 consists of: mark#(f(X, Y)) =#> a!6220!6220f#(mark(X), Y) mark#(f(X, Y)) =#> mark#(X) Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, 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 : 0, 1 This graph has the following strongly connected components: P_2: mark#(f(X, Y)) =#> mark#(X) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_1, R_0, m, f) by (P_2, R_0, m, f).
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Stand 20472