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Higher Order Rewriting Union Beta pair #487093897
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
n150.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.471437 seconds
cpu usage
0.458976
user time
0.387739
system time
0.071237
max virtual memory
113188.0
max residence set size
9476.0
stage attributes
key
value
starexec-result
YES
output
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: || Input TRS: || 1: g(h(g(PeRCenTX))) -> g(PeRCenTX) || 2: g(g(PeRCenTX)) -> g(h(g(PeRCenTX))) || 3: h(h(PeRCenTX)) -> h(f(h(PeRCenTX),PeRCenTX)) || 4: TIlDePAIR(PeRCenTX,PeRCenTY) -> PeRCenTX || 5: TIlDePAIR(PeRCenTX,PeRCenTY) -> PeRCenTY || Number of strict rules: 5 || Direct POLO(bPol) ... failed. || Uncurrying ... failed. || Dependency Pairs: || #1: #g(g(PeRCenTX)) -> #g(h(g(PeRCenTX))) || #2: #g(g(PeRCenTX)) -> #h(g(PeRCenTX)) || #3: #h(h(PeRCenTX)) -> #h(f(h(PeRCenTX),PeRCenTX)) || Number of SCCs: 0, DPs: 0 || We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [FuhKop19]). We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: 0] map#(F, cons(X, Y)) =#> map#(F, Y) 1] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 2] filter2#(true, F, X, Y) =#> filter#(F, Y) 3] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: g(h(g(X))) => g(X) g(g(X)) => g(h(g(X))) h(h(X)) => h(f(h(X), X)) 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) Thus, the original system is terminating if (P_0, R_0, computable, formative) is finite. We consider the dependency pair problem (P_0, R_0, computable, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : 0 * 1 : 2, 3 * 2 : 1 * 3 : 1 This graph has the following strongly connected components: P_1: map#(F, cons(X, Y)) =#> map#(F, Y) P_2: filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) filter2#(true, F, X, Y) =#> filter#(F, Y) filter2#(false, F, X, Y) =#> filter#(F, Y)
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