- : unit = () h : heuristic = - : unit = () APPLY CRITERIA (Marked dependency pairs) TRS termination of: [1] f(x,empty) -> x [2] f(empty,cons(a,k)) -> f(cons(a,k),k) [3] f(cons(a,k),y) -> f(y,k) Sub problem: guided: DP termination of: END GUIDED APPLY CRITERIA (Graph splitting) Found 1 components: { --> --> --> } APPLY CRITERIA (Choosing graph) Trying to solve the following constraints: { f(empty,cons(a,k)) >= f(cons(a,k),k) ; f(cons(a,k),y) >= f(y,k) ; f(x,empty) >= x ; Marked_f(empty,cons(a,k)) >= Marked_f(cons(a,k),k) ; Marked_f(cons(a,k),y) >= Marked_f(y,k) ; } + Disjunctions:{ { Marked_f(empty,cons(a,k)) > Marked_f(cons(a,k),k) ; } { Marked_f(cons(a,k),y) > Marked_f(y,k) ; } } === TIMER virtual : 10.000000 === Entering poly_solver Starting Sat solver initialization Calling Sat solver... === STOPING TIMER virtual === === TIMER real : 10.000000 === === STOPING TIMER real === Sat solver returned Sat solver result read === STOPING TIMER real === === STOPING TIMER virtual === constraint: f(empty,cons(a,k)) >= f(cons(a,k),k) constraint: f(cons(a,k),y) >= f(y,k) constraint: f(x,empty) >= x constraint: Marked_f(empty,cons(a,k)) >= Marked_f(cons(a,k),k) constraint: Marked_f(cons(a,k),y) >= Marked_f(y,k) APPLY CRITERIA (Graph splitting) Found 1 components: { --> } APPLY CRITERIA (Choosing graph) Trying to solve the following constraints: { f(empty,cons(a,k)) >= f(cons(a,k),k) ; f(cons(a,k),y) >= f(y,k) ; f(x,empty) >= x ; Marked_f(cons(a,k),y) >= Marked_f(y,k) ; } + Disjunctions:{ { Marked_f(cons(a,k),y) > Marked_f(y,k) ; } } === TIMER virtual : 10.000000 === Entering poly_solver Starting Sat solver initialization Calling Sat solver... === STOPING TIMER virtual === === TIMER real : 10.000000 === === STOPING TIMER real === Sat solver returned Sat solver result read === STOPING TIMER real === === STOPING TIMER virtual === constraint: f(empty,cons(a,k)) >= f(cons(a,k),k) constraint: f(cons(a,k),y) >= f(y,k) constraint: f(x,empty) >= x constraint: Marked_f(cons(a,k),y) >= Marked_f(y,k) APPLY CRITERIA (Graph splitting) Found 0 components: SOLVED { TRS termination of: [1] f(x,empty) -> x [2] f(empty,cons(a,k)) -> f(cons(a,k),k) [3] f(cons(a,k),y) -> f(y,k) , CRITERION: MDP [ { DP termination of: , CRITERION: SG [ { DP termination of: , CRITERION: CG using polynomial interpretation = [ f ] (X0,X1) = 2*X1 + 1*X0 + 3; [ cons ] (X0,X1) = 2*X1; [ empty ] () = 2; [ Marked_f ] (X0,X1) = 2*X1 + 1*X0; removing [ { DP termination of: , CRITERION: SG [ { DP termination of: , CRITERION: ORD [ Solution found: polynomial interpretation = [ f ] (X0,X1) = 3 + 1*X0 + 2*X1 + 0; [ cons ] (X0,X1) = 3 + 1*X0 + 3*X1 + 0; [ empty ] () = 2 + 0; [ Marked_f ] (X0,X1) = 2*X0 + 3*X1 + 0; ]} ]} ]} ]} ]} Cime worked for 0.029794 seconds (real time) Cime Exit Status: 0