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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270715
details
property
value
status
complete
benchmark
Applicative_first_order_05__hydra.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n066.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
1.06626 seconds
cpu usage
1.9975
user time
1.86378
system time
0.133724
max virtual memory
6956436.0
max residence set size
116772.0
stage attributes
key
value
starexec-result
YES
output
1.91/1.05 YES 1.91/1.06 We consider the system theBenchmark. 1.91/1.06 1.91/1.06 Alphabet: 1.91/1.06 1.91/1.06 0 : [] --> a 1.91/1.06 cons : [c * c] --> c 1.91/1.06 copy : [a * c * c] --> c 1.91/1.06 f : [c] --> c 1.91/1.06 false : [] --> b 1.91/1.06 filter : [c -> b * c] --> c 1.91/1.06 filter2 : [b * c -> b * c * c] --> c 1.91/1.06 map : [c -> c * c] --> c 1.91/1.06 n : [] --> a 1.91/1.06 nil : [] --> c 1.91/1.06 s : [a] --> a 1.91/1.06 true : [] --> b 1.91/1.06 1.91/1.06 Rules: 1.91/1.06 1.91/1.06 f(cons(nil, x)) => x 1.91/1.06 f(cons(f(cons(nil, x)), y)) => copy(n, x, y) 1.91/1.06 copy(0, x, y) => f(y) 1.91/1.06 copy(s(x), y, z) => copy(x, y, cons(f(y), z)) 1.91/1.06 map(g, nil) => nil 1.91/1.06 map(g, cons(x, y)) => cons(g x, map(g, y)) 1.91/1.06 filter(g, nil) => nil 1.91/1.06 filter(g, cons(x, y)) => filter2(g x, g, x, y) 1.91/1.06 filter2(true, g, x, y) => cons(x, filter(g, y)) 1.91/1.06 filter2(false, g, x, y) => filter(g, y) 1.91/1.06 1.91/1.06 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 1.91/1.06 1.91/1.06 We observe that the rules contain a first-order subset: 1.91/1.06 1.91/1.06 f(cons(nil, X)) => X 1.91/1.06 f(cons(f(cons(nil, X)), Y)) => copy(n, X, Y) 1.91/1.06 copy(0, X, Y) => f(Y) 1.91/1.06 copy(s(X), Y, Z) => copy(X, Y, cons(f(Y), Z)) 1.91/1.06 1.91/1.06 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. 1.91/1.06 1.91/1.06 According to the external first-order termination prover, this system is indeed Ce-terminating: 1.91/1.06 1.91/1.06 || proof of resources/system.trs 1.91/1.06 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 1.91/1.06 || 1.91/1.06 || 1.91/1.06 || Termination w.r.t. Q of the given QTRS could be proven: 1.91/1.06 || 1.91/1.06 || (0) QTRS 1.91/1.06 || (1) DependencyPairsProof [EQUIVALENT] 1.91/1.06 || (2) QDP 1.91/1.06 || (3) DependencyGraphProof [EQUIVALENT] 1.91/1.06 || (4) QDP 1.91/1.06 || (5) QDPSizeChangeProof [EQUIVALENT] 1.91/1.06 || (6) YES 1.91/1.06 || 1.91/1.06 || 1.91/1.06 || ---------------------------------------- 1.91/1.06 || 1.91/1.06 || (0) 1.91/1.06 || Obligation: 1.91/1.06 || Q restricted rewrite system: 1.91/1.06 || The TRS R consists of the following rules: 1.91/1.06 || 1.91/1.06 || f(cons(nil, %X)) -> %X 1.91/1.06 || f(cons(f(cons(nil, %X)), %Y)) -> copy(n, %X, %Y) 1.91/1.06 || copy(0, %X, %Y) -> f(%Y) 1.91/1.06 || copy(s(%X), %Y, %Z) -> copy(%X, %Y, cons(f(%Y), %Z)) 1.91/1.06 || ~PAIR(%X, %Y) -> %X 1.91/1.06 || ~PAIR(%X, %Y) -> %Y 1.91/1.06 || 1.91/1.06 || Q is empty. 1.91/1.06 || 1.91/1.06 || ---------------------------------------- 1.91/1.06 || 1.91/1.06 || (1) DependencyPairsProof (EQUIVALENT) 1.91/1.06 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. 1.91/1.06 || ---------------------------------------- 1.91/1.06 || 1.91/1.06 || (2) 1.91/1.06 || Obligation: 1.91/1.06 || Q DP problem: 1.91/1.06 || The TRS P consists of the following rules: 1.91/1.06 || 1.91/1.06 || F(cons(f(cons(nil, %X)), %Y)) -> COPY(n, %X, %Y) 1.91/1.06 || COPY(0, %X, %Y) -> F(%Y) 1.91/1.06 || COPY(s(%X), %Y, %Z) -> COPY(%X, %Y, cons(f(%Y), %Z)) 1.91/1.06 || COPY(s(%X), %Y, %Z) -> F(%Y) 1.91/1.06 || 1.91/1.06 || The TRS R consists of the following rules: 1.91/1.06 || 1.91/1.06 || f(cons(nil, %X)) -> %X 1.91/1.06 || f(cons(f(cons(nil, %X)), %Y)) -> copy(n, %X, %Y) 1.91/1.06 || copy(0, %X, %Y) -> f(%Y) 1.91/1.06 || copy(s(%X), %Y, %Z) -> copy(%X, %Y, cons(f(%Y), %Z)) 1.91/1.06 || ~PAIR(%X, %Y) -> %X 1.91/1.06 || ~PAIR(%X, %Y) -> %Y 1.91/1.06 ||
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