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Higher Order Rewriting Union Beta pair #487093873
details
property
value
status
complete
benchmark
qsort.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n138.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
1.63606 seconds
cpu usage
1.6181
user time
1.47456
system time
0.143541
max virtual memory
146456.0
max residence set size
32308.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat app : [list * list] --> list cons : [nat * list] --> list false : [] --> bool filter : [nat -> bool * list] --> list gr : [nat * nat] --> bool if : [bool * list * list] --> list le : [nat * nat] --> bool nil : [] --> list qsort : [list] --> list s : [nat] --> nat true : [] --> bool Rules: if(true, x, y) => x if(false, x, y) => y app(nil, x) => x app(cons(x, y), z) => cons(x, app(y, z)) le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) gr(0, x) => false gr(s(x), 0) => true gr(s(x), s(y)) => gr(x, y) filter(f, nil) => nil filter(f, cons(x, y)) => if(f x, cons(x, filter(f, y)), filter(f, y)) qsort(nil) => nil qsort(cons(x, y)) => app(qsort(filter(/\z.le(z, x), y)), app(cons(x, nil), qsort(filter(/\u.gr(u, x), 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: if(true, X, Y) => X if(false, X, Y) => Y app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) gr(0, X) => false gr(s(X), 0) => true gr(s(X), s(Y)) => gr(X, 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: if(true(),PeRCenTX,PeRCenTY) -> PeRCenTX || 2: if(false(),PeRCenTX,PeRCenTY) -> PeRCenTY || 3: app(nil(),PeRCenTX) -> PeRCenTX || 4: app(cons(PeRCenTX,PeRCenTY),PeRCenTZ) -> cons(PeRCenTX,app(PeRCenTY,PeRCenTZ)) || 5: le(0(),PeRCenTX) -> true() || 6: le(s(PeRCenTX),0()) -> false() || 7: le(s(PeRCenTX),s(PeRCenTY)) -> le(PeRCenTX,PeRCenTY) || 8: gr(0(),PeRCenTX) -> false() || 9: gr(s(PeRCenTX),0()) -> true() || 10: gr(s(PeRCenTX),s(PeRCenTY)) -> gr(PeRCenTX,PeRCenTY) || Number of strict rules: 10 || Direct POLO(bPol) ... removes: 4 8 1 3 5 10 7 9 6 2 || le w: 2 * x1 + 2 * x2 || s w: 2 * x1 + 1 || false w: 1 || true w: 1 || if w: 2 * x1 + 2 * x2 + 2 * x3 || 0 w: 1 || nil w: 1 || cons w: x1 + x2 + 1 || gr w: 2 * x1 + x2 || app w: 2 * x1 + 2 * x2 + 1 || Number of strict rules: 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] filter#(F, cons(X, Y)) =#> if#(F X, cons(X, filter(F, Y)), filter(F, Y)) 1] filter#(F, cons(X, Y)) =#> filter#(F, Y) 2] filter#(F, cons(X, Y)) =#> filter#(F, Y) 3] qsort#(cons(X, Y)) =#> app#(qsort(filter(/\x.le(x, X), Y)), app(cons(X, nil), qsort(filter(/\y.gr(y, X), Y)))) 4] qsort#(cons(X, Y)) =#> qsort#(filter(/\x.le(x, X), Y)) 5] qsort#(cons(X, Y)) =#> filter#(/\x.le(x, X), Y) 6] qsort#(cons(X, Y)) =#> le#(Z, X) 7] qsort#(cons(X, Y)) =#> app#(cons(X, nil), qsort(filter(/\x.gr(x, X), Y))) 8] qsort#(cons(X, Y)) =#> qsort#(filter(/\x.gr(x, X), Y)) 9] qsort#(cons(X, Y)) =#> filter#(/\x.gr(x, X), Y) 10] qsort#(cons(X, Y)) =#> gr#(Z, X) Rules R_0: if(true, X, Y) => X
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