DONT KNOW Welcome to CiME version 2.02 - Built on 14/04/2004 13:29:51 - : unit = () - : unit = () X : variable_set = F : signature = R : HTRS = { zeros -> cons(0,zeros), U11(tt,V1) -> U12(isNatList(V1)), U12(tt) -> tt, U21(tt,V1) -> U22(isNat(V1)), U22(tt) -> tt, U31(tt,V) -> U32(isNatList(V)), U32(tt) -> tt, U41(tt,V1,V2) -> U42(isNat(V1),V2), U42(tt,V2) -> U43(isNatIList(V2)), U43(tt) -> tt, U51(tt,V1,V2) -> U52(isNat(V1),V2), U52(tt,V2) -> U53(isNatList(V2)), U53(tt) -> tt, U61(tt,V1,V2) -> U62(isNat(V1),V2), U62(tt,V2) -> U63(isNatIList(V2)), U63(tt) -> tt, U71(tt,L) -> s(length(L)), U81(tt) -> nil, U91(tt,IL,M,N) -> cons(N,take(M,IL)), and(tt,X) -> X, isNat(0) -> tt, isNat(length(V1)) -> U11(isNatIListKind(V1),V1), isNat(s(V1)) -> U21(isNatKind(V1),V1), isNatIList(V) -> U31(isNatIListKind(V),V), isNatIList(zeros) -> tt, isNatIList(cons(V1,V2)) -> U41(and(isNatKind(V1),isNatIListKind(V2)),V1,V2), isNatIListKind(nil) -> tt, isNatIListKind(zeros) -> tt, isNatIListKind(cons(V1,V2)) -> and(isNatKind(V1),isNatIListKind(V2)), isNatIListKind(take(V1,V2)) -> and(isNatKind(V1),isNatIListKind(V2)), isNatKind(0) -> tt, isNatKind(length(V1)) -> isNatIListKind(V1), isNatKind(s(V1)) -> isNatKind(V1), isNatList(nil) -> tt, isNatList(cons(V1,V2)) -> U51(and(isNatKind(V1),isNatIListKind(V2)),V1,V2), isNatList(take(V1,V2)) -> U61(and(isNatKind(V1),isNatIListKind(V2)),V1,V2), length(nil) -> 0, length(cons(N,L)) -> U71(and(and(isNatList(L),isNatIListKind(L)),and(isNat(N), isNatKind(N))),L), take(0,IL) -> U81(and(isNatIList(IL),isNatIListKind(IL))), take(s(M),cons(N,IL)) -> U91(and(and(isNatIList(IL),isNatIListKind(IL)),and(and(isNat(M), isNatKind(M)), and(isNat(N), isNatKind(N)))),IL,M,N) } (40 rules) Termination now uses minimal decomposition - : unit = () Entering the termination expert for modules. Verbose level = 0 Checking module: {} The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: { and(tt,V_7) -> V_7 } (1 rules) The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: {} The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: {} The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: { U81(tt) -> nil } (1 rules) The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: {} The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: { U43(tt) -> tt } (1 rules) The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: { U63(tt) -> tt } (1 rules) The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: { U53(tt) -> tt } (1 rules) The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: {} The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: { U22(tt) -> tt } (1 rules) The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: { U12(tt) -> tt } (1 rules) The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: { U32(tt) -> tt } (1 rules) The dependency graph is (0 nodes) Checking each of the 0 strongly connected components : Checking module: { zeros -> cons(0,zeros) } (1 rules) The dependency graph is (1 nodes) Checking each of the 1 strongly connected components : Checking component 1 Trying simple graph criterion. Trying to solve the following constraints: (2 termination constraints) Search parameters: linear polynomials, coefficient bound is 2. No solution found for these parameters. Search parameters: simple polynomials, coefficient bound is 3. No solution found for these parameters. No solution found for these constraints. Trying strongly connected part criterion. Trying to solve the following constraints: (2 termination constraints) Search parameters: linear polynomials, coefficient bound is 2. No solution found for these parameters. Search parameters: simple polynomials, coefficient bound is 3. No solution found for these parameters. No solution found for these constraints. No modular termination proof found. - : unit = () The last proof attempt failed. - : unit = () Quitting. Standard error: