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HRS union beta 16688 pair #381734869
details
property
value
status
complete
benchmark
Applicative_first_order_05__#3.40.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n041.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
1.22202992439 seconds
cpu usage
2.867324843
max memory
1.68067072E8
stage attributes
key
value
output-size
20379
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 : [] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d minus : [a * a] --> a nil : [] --> d plus : [a * a] --> a quot : [a * a] --> a s : [a] --> a true : [] --> b Rules: minus(x, 0) => x minus(s(x), s(y)) => minus(x, y) quot(0, s(x)) => 0 quot(s(x), s(y)) => s(quot(minus(x, y), s(y))) plus(0, x) => x plus(s(x), y) => s(plus(x, y)) plus(minus(x, s(0)), minus(y, s(s(z)))) => plus(minus(y, s(s(z))), minus(x, s(0))) plus(plus(x, s(0)), plus(y, s(s(z)))) => plus(plus(y, s(s(z))), plus(x, s(0))) 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: minus(X, 0) => X minus(s(X), s(Y)) => minus(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(minus(X, Y), s(Y))) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) plus(minus(X, s(0)), minus(Y, s(s(Z)))) => plus(minus(Y, s(s(Z))), minus(X, s(0))) plus(plus(X, s(0)), plus(Y, s(s(Z)))) => plus(plus(Y, s(s(Z))), plus(X, s(0))) 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) AND || (5) QDP || (6) UsableRulesProof [EQUIVALENT] || (7) QDP || (8) MRRProof [EQUIVALENT] || (9) QDP || (10) MRRProof [EQUIVALENT] || (11) QDP || (12) MRRProof [EQUIVALENT] || (13) QDP || (14) DependencyGraphProof [EQUIVALENT] || (15) TRUE || (16) QDP || (17) UsableRulesProof [EQUIVALENT] || (18) QDP || (19) QDPSizeChangeProof [EQUIVALENT] || (20) YES || (21) QDP || (22) QDPOrderProof [EQUIVALENT] || (23) QDP || (24) PisEmptyProof [EQUIVALENT] || (25) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || minus(%X, 0) -> %X
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