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TRS Stand 20472 pair #381710908
details
property
value
status
complete
benchmark
20.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n079.star.cs.uiowa.edu
space
Der95
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.125669002533 seconds
cpu usage
0.115902401
max memory
5468160.0
stage attributes
key
value
output-size
4484
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. and : [o * o] --> o not : [o] --> o or : [o * o] --> o not(not(X)) => X not(or(X, Y)) => and(not(not(not(X))), not(not(not(Y)))) not(and(X, Y)) => or(not(not(not(X))), not(not(not(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] not#(or(X, Y)) =#> not#(not(not(X))) 1] not#(or(X, Y)) =#> not#(not(X)) 2] not#(or(X, Y)) =#> not#(X) 3] not#(or(X, Y)) =#> not#(not(not(Y))) 4] not#(or(X, Y)) =#> not#(not(Y)) 5] not#(or(X, Y)) =#> not#(Y) 6] not#(and(X, Y)) =#> not#(not(not(X))) 7] not#(and(X, Y)) =#> not#(not(X)) 8] not#(and(X, Y)) =#> not#(X) 9] not#(and(X, Y)) =#> not#(not(not(Y))) 10] not#(and(X, Y)) =#> not#(not(Y)) 11] not#(and(X, Y)) =#> not#(Y) Rules R_0: not(not(X)) => X not(or(X, Y)) => and(not(not(not(X))), not(not(not(Y)))) not(and(X, Y)) => or(not(not(not(X))), not(not(not(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: not#(or(X, Y)) >? not#(not(not(X))) not#(or(X, Y)) >? not#(not(X)) not#(or(X, Y)) >? not#(X) not#(or(X, Y)) >? not#(not(not(Y))) not#(or(X, Y)) >? not#(not(Y)) not#(or(X, Y)) >? not#(Y) not#(and(X, Y)) >? not#(not(not(X))) not#(and(X, Y)) >? not#(not(X)) not#(and(X, Y)) >? not#(X) not#(and(X, Y)) >? not#(not(not(Y))) not#(and(X, Y)) >? not#(not(Y)) not#(and(X, Y)) >? not#(Y) not(not(X)) >= X not(or(X, Y)) >= and(not(not(not(X))), not(not(not(Y)))) not(and(X, Y)) >= or(not(not(not(X))), not(not(not(Y)))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: and = \y0y1.1 + y1 + 2y0 not = \y0.y0 not# = \y0.2y0 or = \y0y1.1 + y1 + 2y0 Using this interpretation, the requirements translate to: [[not#(or(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x0 = [[not#(not(not(_x0)))]] [[not#(or(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x0 = [[not#(not(_x0))]] [[not#(or(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x0 = [[not#(_x0)]] [[not#(or(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x1 = [[not#(not(not(_x1)))]] [[not#(or(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x1 = [[not#(not(_x1))]] [[not#(or(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x1 = [[not#(_x1)]] [[not#(and(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x0 = [[not#(not(not(_x0)))]] [[not#(and(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x0 = [[not#(not(_x0))]] [[not#(and(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x0 = [[not#(_x0)]] [[not#(and(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x1 = [[not#(not(not(_x1)))]] [[not#(and(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x1 = [[not#(not(_x1))]] [[not#(and(_x0, _x1))]] = 2 + 2x1 + 4x0 > 2x1 = [[not#(_x1)]] [[not(not(_x0))]] = x0 >= x0 = [[_x0]] [[not(or(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[and(not(not(not(_x0))), not(not(not(_x1))))]] [[not(and(_x0, _x1))]] = 1 + x1 + 2x0 >= 1 + x1 + 2x0 = [[or(not(not(not(_x0))), not(not(not(_x1))))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_0) by ({}, R_0). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. +++ Citations +++
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