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Higher Order Rewriting Union Beta pair #487093949
details
property
value
status
complete
benchmark
Applicative_first_order_05__13.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
107.135 seconds
cpu usage
383.975
user time
380.881
system time
3.09357
max virtual memory
5166704.0
max residence set size
1700480.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: !facplus : [a * a] --> a !factimes : [a * a] --> 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 true : [] --> b Rules: !factimes(x, !facplus(y, z)) => !facplus(!factimes(x, y), !factimes(x, z)) !factimes(!facplus(x, y), z) => !facplus(!factimes(z, x), !factimes(z, y)) !factimes(!factimes(x, y), z) => !factimes(x, !factimes(y, z)) !facplus(!facplus(x, y), z) => !facplus(x, !facplus(y, z)) 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 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] !factimes#(X, !facplus(Y, Z)) =#> !facplus#(!factimes(X, Y), !factimes(X, Z)) 1] !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Y) 2] !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Z) 3] !factimes#(!facplus(X, Y), Z) =#> !facplus#(!factimes(Z, X), !factimes(Z, Y)) 4] !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, X) 5] !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, Y) 6] !factimes#(!factimes(X, Y), Z) =#> !factimes#(X, !factimes(Y, Z)) 7] !factimes#(!factimes(X, Y), Z) =#> !factimes#(Y, Z) 8] !facplus#(!facplus(X, Y), Z) =#> !facplus#(X, !facplus(Y, Z)) 9] !facplus#(!facplus(X, Y), Z) =#> !facplus#(Y, Z) 10] map#(F, cons(X, Y)) =#> map#(F, Y) 11] filter#(F, cons(X, Y)) =#> filter2#(F X, F, X, Y) 12] filter2#(true, F, X, Y) =#> filter#(F, Y) 13] filter2#(false, F, X, Y) =#> filter#(F, Y) Rules R_0: !factimes(X, !facplus(Y, Z)) => !facplus(!factimes(X, Y), !factimes(X, Z)) !factimes(!facplus(X, Y), Z) => !facplus(!factimes(Z, X), !factimes(Z, Y)) !factimes(!factimes(X, Y), Z) => !factimes(X, !factimes(Y, Z)) !facplus(!facplus(X, Y), Z) => !facplus(X, !facplus(Y, Z)) 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 : 8, 9 * 1 : 0, 1, 2, 3, 4, 5, 6, 7 * 2 : 0, 1, 2, 3, 4, 5, 6, 7 * 3 : 8, 9 * 4 : 0, 1, 2, 3, 4, 5, 6, 7 * 5 : 0, 1, 2, 3, 4, 5, 6, 7 * 6 : 0, 1, 2, 3, 4, 5, 6, 7 * 7 : 0, 1, 2, 3, 4, 5, 6, 7 * 8 : 8, 9 * 9 : 8, 9 * 10 : 10 * 11 : 12, 13 * 12 : 11 * 13 : 11 This graph has the following strongly connected components: P_1: !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Y) !factimes#(X, !facplus(Y, Z)) =#> !factimes#(X, Z) !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, X) !factimes#(!facplus(X, Y), Z) =#> !factimes#(Z, Y) !factimes#(!factimes(X, Y), Z) =#> !factimes#(X, !factimes(Y, Z)) !factimes#(!factimes(X, Y), Z) =#> !factimes#(Y, Z) P_2: !facplus#(!facplus(X, Y), Z) =#> !facplus#(X, !facplus(Y, Z)) !facplus#(!facplus(X, Y), Z) =#> !facplus#(Y, Z)
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