Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
HRS union beta 16688 pair #381734878
details
property
value
status
complete
benchmark
Applicative_first_order_05__#3.52.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n105.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.778715133667 seconds
cpu usage
1.523579244
max memory
7.2380416E7
stage attributes
key
value
output-size
7277
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: 0 : [] --> a 1 : [] --> a cons : [c * d] --> d f : [a * a * a] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d nil : [] --> d s : [a] --> a true : [] --> b Rules: f(0, 1, x) => f(s(x), x, x) f(x, y, s(z)) => s(f(0, 1, z)) map(g, nil) => nil map(g, cons(x, y)) => cons(g x, map(g, y)) filter(g, nil) => nil filter(g, cons(x, y)) => filter2(g x, g, x, y) filter2(true, g, x, y) => cons(x, filter(g, y)) filter2(false, g, x, y) => filter(g, 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: f(0, 1, X) => f(s(X), X, X) f(X, Y, s(Z)) => s(f(0, 1, Z)) Moreover, the system is finitely branching. 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 Ce-terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed Ce-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) DependencyPairsProof [EQUIVALENT] || (2) QDP || (3) QDPSizeChangeProof [EQUIVALENT] || (4) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || f(0, 1, %X) -> f(s(%X), %X, %X) || f(%X, %Y, s(%Z)) -> s(f(0, 1, %Z)) || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || || ---------------------------------------- || || (1) DependencyPairsProof (EQUIVALENT) || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. || ---------------------------------------- || || (2) || Obligation: || Q DP problem: || The TRS P consists of the following rules: || || F(0, 1, %X) -> F(s(%X), %X, %X) || F(%X, %Y, s(%Z)) -> F(0, 1, %Z) || || The TRS R consists of the following rules: || || f(0, 1, %X) -> f(s(%X), %X, %X) || f(%X, %Y, s(%Z)) -> s(f(0, 1, %Z)) || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || We have to consider all minimal (P,Q,R)-chains. || ---------------------------------------- || || (3) QDPSizeChangeProof (EQUIVALENT) || By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. ||
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to HRS union beta 16688