+++ /dev/null
-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 *)