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HRS union beta 16688 pair #381734941
details
property
value
status
complete
benchmark
Applicative_first_order_05__motivation.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
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.866266965866 seconds
cpu usage
1.830779582
max memory
1.18419456E8
stage attributes
key
value
output-size
8714
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: cons : [c * d] --> d f : [a * a] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d g : [a] --> a h : [a] --> a map : [c -> c * d] --> d nil : [] --> d true : [] --> b Rules: g(h(g(x))) => g(x) g(g(x)) => g(h(g(x))) h(h(x)) => h(f(h(x), x)) map(i, nil) => nil map(i, cons(x, y)) => cons(i x, map(i, y)) filter(i, nil) => nil filter(i, cons(x, y)) => filter2(i x, i, x, y) filter2(true, i, x, y) => cons(x, filter(i, y)) filter2(false, i, x, y) => filter(i, 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: g(h(g(X))) => g(X) g(g(X)) => g(h(g(X))) h(h(X)) => h(f(h(X), X)) 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) DependencyGraphProof [EQUIVALENT] || (4) QDP || (5) QDPOrderProof [EQUIVALENT] || (6) QDP || (7) PisEmptyProof [EQUIVALENT] || (8) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || g(h(g(%X))) -> g(%X) || g(g(%X)) -> g(h(g(%X))) || h(h(%X)) -> h(f(h(%X), %X)) || ~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: || || G(g(%X)) -> G(h(g(%X))) || G(g(%X)) -> H(g(%X)) || H(h(%X)) -> H(f(h(%X), %X)) || || The TRS R consists of the following rules: || || g(h(g(%X))) -> g(%X) || g(g(%X)) -> g(h(g(%X))) || h(h(%X)) -> h(f(h(%X), %X)) || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y
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