ceta_eq: termination proof not accepted 1: error below switch to dependency pairs 1.1: error below the dependency graph processor 1.1.1: error when applying the reduction pair processor with usable rules to remove from the DP problem pairs: quicksort#(cons(N, L)) -> quicksort#(low(N, L)) quicksort#(cons(N, L)) -> quicksort#(high(N, L)) rules: app(nil, Y) -> Y app(cons(N, L), Y) -> cons(N, app(L, Y)) high(N, nil) -> nil high(N, cons(M, L)) -> ifhigh(le(M, N), N, cons(M, L)) ifhigh(true, N, cons(M, L)) -> high(N, L) ifhigh(false, N, cons(M, L)) -> cons(M, high(N, L)) iflow(true, N, cons(M, L)) -> cons(M, low(N, L)) iflow(false, N, cons(M, L)) -> low(N, L) le(0, Y) -> true le(s(X), 0) -> false le(s(X), s(Y)) -> le(X, Y) low(N, nil) -> nil low(N, cons(M, L)) -> iflow(le(M, N), N, cons(M, L)) quicksort(nil) -> nil quicksort(cons(N, L)) -> app(quicksort(low(N, L)), cons(N, quicksort(high(N, L)))) the pairs quicksort#(cons(N, L)) -> quicksort#(low(N, L)) could not apply the generic root reduction pair processor with the following SCNP-version with mu = MAX and the level mapping defined by pi(quicksort#) = [(epsilon,0),(1,1)] Argument Filter: pi(quicksort#/1) = 1 pi(cons/2) = [2] pi(high/2) = [2] pi(low/2) = 2 pi(nil/0) = [] pi(ifhigh/3) = [3] pi(le/2) = [1,2] pi(true/0) = [] pi(iflow/3) = 3 pi(false/0) = [] pi(0/0) = [] pi(s/1) = [] RPO with the following precedence precedence(cons[2]) = 1 precedence(high[2]) = 1 precedence(nil[0]) = 1 precedence(ifhigh[3]) = 1 precedence(le[2]) = 0 precedence(true[0]) = 2 precedence(false[0]) = 0 precedence(0[0]) = 0 precedence(s[1]) = 0 precedence(_) = 0 and the following status status(cons[2]) = mul status(high[2]) = mul status(nil[0]) = mul status(ifhigh[3]) = mul status(le[2]) = mul status(true[0]) = mul status(false[0]) = mul status(0[0]) = mul status(s[1]) = mul status(_) = lex problem when orienting (usable) rules could not orient ifhigh(false, N, cons(M, L)) >= cons(M, high(N, L)) pi( ifhigh(false, N, cons(M, L)) ) = ifhigh(cons(L)) pi( cons(M, high(N, L)) ) = cons(high(L)) could not orient ifhigh(cons(L)) >=RPO cons(high(L))