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Higher Order Rewriting Union Beta pair #487093951
details
property
value
status
complete
benchmark
Applicative_AG01_innermost__#4.22.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n137.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.989758 seconds
cpu usage
0.394279
user time
0.337159
system time
0.05712
max virtual memory
125672.0
max residence set size
10392.0
stage attributes
key
value
starexec-result
YES
output
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 nil : [] --> d quot : [a * a * a] --> a s : [a] --> a true : [] --> b Rules: quot(0, s(x), s(y)) => 0 quot(s(x), s(y), z) => quot(x, y, z) quot(x, 0, s(y)) => s(quot(x, s(y), s(y))) 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: quot(0, s(X), s(Y)) => 0 quot(s(X), s(Y), Z) => quot(X, Y, Z) quot(X, 0, s(Y)) => s(quot(X, s(Y), s(Y))) Moreover, the system is orthogonal. 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 terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed terminating: || Input TRS: || 1: quot(0(),s(PeRCenTX),s(PeRCenTY)) -> 0() || 2: quot(s(PeRCenTX),s(PeRCenTY),PeRCenTZ) -> quot(PeRCenTX,PeRCenTY,PeRCenTZ) || 3: quot(PeRCenTX,0(),s(PeRCenTY)) -> s(quot(PeRCenTX,s(PeRCenTY),s(PeRCenTY))) || Number of strict rules: 3 || Direct POLO(bPol) ... failed. || Uncurrying ... failed. || Dependency Pairs: || #1: #quot(s(PeRCenTX),s(PeRCenTY),PeRCenTZ) -> #quot(PeRCenTX,PeRCenTY,PeRCenTZ) || #2: #quot(PeRCenTX,0(),s(PeRCenTY)) -> #quot(PeRCenTX,s(PeRCenTY),s(PeRCenTY)) || Number of SCCs: 1, DPs: 2 || SCC { #1 #2 } || POLO(Sum)... succeeded. || s w: x1 + 1 || 0 w: 1 || quot w: 0 || #quot w: x1 || USABLE RULES: { } || Removed DPs: #1 || 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: quot(0, s(X), s(Y)) => 0 quot(s(X), s(Y), Z) => quot(X, Y, Z) quot(X, 0, s(Y)) => s(quot(X, s(Y), s(Y))) 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)
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