alias nat          /Coq/Init/Datatypes/nat.ind#1/1
alias eqT          /Coq/Init/Logic_Type/eqT.ind#1/1
alias eq           /Coq/Init/Logic/Equality/eq.ind#1/1
alias refl_equal   /Coq/Init/Logic/Equality/eq.ind#1/1/1
alias eq_ind       /Coq/Init/Logic/Equality/eq_ind.con
alias eq_ind_r     /Coq/Init/Logic/Logic_lemmas/eq_ind_r.con
alias O            /Coq/Init/Datatypes/nat.ind#1/1/1
alias S            /Coq/Init/Datatypes/nat.ind#1/1/2
alias plus         /Coq/Init/Peano/plus.con
alias mult         /Coq/Init/Peano/mult.con
alias le           /Coq/Init/Peano/le.ind#1/1
alias lt           /Coq/Init/Peano/lt.con
alias not          /Coq/Init/Logic/not.con
alias f_equal      /Coq/Init/Logic/Logic_lemmas/equality/f_equal.con
alias le_trans     /Coq/Arith/Le/le_trans.con

alias le_plus_plus /Coq/Arith/Plus/le_plus_plus.con
alias le_reg_r     /Coq/Arith/Plus/le_reg_r.con
alias le_reg_l     /Coq/Arith/Plus/le_reg_l.con

alias plus_n_O     /Coq/Init/Peano/plus_n_O.con 

!n:nat.!m:nat.(le n m)->(le (mult (S (S O)) n) (mult (S (S O)) m))

(* Lo scopo dell'esercizio e' riuscire a effettuare la dimostrazione che *)
(* (n <= m) -> (2*n <= 2*m) come la si farebbe su carta, ovvero:         *)
(*                                                                       *)
(*     2 * n                                                             *)
(*  == n + n + 0     Simpl                                               *)
(*  <= m + n + 0     le_reg_r because n <= m because hypothesis          *)
(*  <= m + m + 0     le_reg_l because n + 0 <= m + 0 because le_reg_r    *)
(*                    because hypothesis                                 *)
(*  == 2 * m         Change                                              *)