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HRS union beta 16688 pair #381734824
details
property
value
status
complete
benchmark
Applicative_AG01_innermost__#4.36.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n087.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
1.38873386383 seconds
cpu usage
3.353096517
max memory
2.12770816E8
stage attributes
key
value
output-size
45284
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: 0 : [] --> b cons : [b * c] --> c eq : [b * b] --> a false : [] --> a filter : [b -> a * c] --> c filter2 : [a * b -> a * b * c] --> c if!fac6220min : [a * c] --> b if!fac6220replace : [a * b * b * c] --> c le : [b * b] --> a map : [b -> b * c] --> c min : [c] --> b nil : [] --> c replace : [b * b * c] --> c s : [b] --> b sort : [c] --> c true : [] --> a Rules: eq(0, 0) => true eq(0, s(x)) => false eq(s(x), 0) => false eq(s(x), s(y)) => eq(x, y) le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) min(cons(0, nil)) => 0 min(cons(s(x), nil)) => s(x) min(cons(x, cons(y, z))) => if!fac6220min(le(x, y), cons(x, cons(y, z))) if!fac6220min(true, cons(x, cons(y, z))) => min(cons(x, z)) if!fac6220min(false, cons(x, cons(y, z))) => min(cons(y, z)) replace(x, y, nil) => nil replace(x, y, cons(z, u)) => if!fac6220replace(eq(x, z), x, y, cons(z, u)) if!fac6220replace(true, x, y, cons(z, u)) => cons(y, u) if!fac6220replace(false, x, y, cons(z, u)) => cons(z, replace(x, y, u)) sort(nil) => nil sort(cons(x, y)) => cons(min(cons(x, y)), sort(replace(min(cons(x, y)), x, y))) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) filter(f, nil) => nil filter(f, cons(x, y)) => filter2(f x, f, x, y) filter2(true, f, x, y) => cons(x, filter(f, y)) filter2(false, f, x, y) => filter(f, y) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We observe that the rules contain a first-order subset: eq(0, 0) => true eq(0, s(X)) => false eq(s(X), 0) => false eq(s(X), s(Y)) => eq(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) min(cons(0, nil)) => 0 min(cons(s(X), nil)) => s(X) min(cons(X, cons(Y, Z))) => if!fac6220min(le(X, Y), cons(X, cons(Y, Z))) if!fac6220min(true, cons(X, cons(Y, Z))) => min(cons(X, Z)) if!fac6220min(false, cons(X, cons(Y, Z))) => min(cons(Y, Z)) replace(X, Y, nil) => nil replace(X, Y, cons(Z, U)) => if!fac6220replace(eq(X, Z), X, Y, cons(Z, U)) if!fac6220replace(true, X, Y, cons(Z, U)) => cons(Y, U) if!fac6220replace(false, X, Y, cons(Z, U)) => cons(Z, replace(X, Y, U)) sort(nil) => nil sort(cons(X, Y)) => cons(min(cons(X, Y)), sort(replace(min(cons(X, Y)), X, Y))) Moreover, the system is orthogonal. Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed terminating: || proof of resources/system.trs || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 || || || Termination w.r.t. Q of the given QTRS could be proven: || || (0) QTRS || (1) Overlay + Local Confluence [EQUIVALENT] || (2) QTRS || (3) DependencyPairsProof [EQUIVALENT] || (4) QDP || (5) DependencyGraphProof [EQUIVALENT] || (6) AND || (7) QDP || (8) UsableRulesProof [EQUIVALENT] || (9) QDP || (10) QReductionProof [EQUIVALENT] || (11) QDP
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