proposition: Lemma contradict_G_prefix {v1 v2 n} : §0^G ++ v1 = §0^n ++ [§1] ++ v2 -> n < G -> False.
proof: Proof. move=> /esym /zero_prefix_lt. by lia. Qed.
imports: Relation_Operators Operators_Properties Lia PeanoNat List Undecidability.StackMachines.Reductions.CM1_HALT_to_SMNdl_UB.SMX_facts Undecidability.StackMachines.Util(Nat_facts List_facts) ssreflect ssrbool ssrfun