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@. Mbdy·tr, T. , ?. ·. Ta, and T. , ? If m ? ?TA, then there is an unique TA i ? TA where altlook(m, TA i ) = fail. ? If m ? ?TA, then, since C i is well-typed, the rule (Cla·Ok) enforces that m ? ?TA. Then, for all TA i ? TA, we have m ? meth(TA i ) ? m in TA i n in TA 1 . Moreover, we know that n ? meth(TA 1 ), by Lemma 1, which means that there is at least one altlook(n, TA 1 ) = I n (I x){. . .} which is derivable, by Lemma 4. Thus, altlook(m, TA i ) = I m (I x){. . .} is derivable, for all TA i such that m ? meth i ) is a function, Which ensures they are all equal

T. If, S. Ok, T. Then-msigm, . If, and S. Only, Proof Straightforward induction on the derivation of msig(m, TA) = S. Here are the most relevant cases: ? (MSig·Ali 2 ) In both rules (Alias·Ok 1 ) and (Alias·Ok 2 ), it is easy to observe that the new name of the method is not in the requires list (the role of (Alias·Ok 2 ) is actually to remove it from the requires list if it appears in the previous step) Hence the result, Lemma, vol.8

T. If and . Ok, )) = I ? I, then mtype(m, TA) = I ? I

@. Ta, Since TA typechecks, then altlook(m, TA 1 ) = fail, by Lemma 4 We also have, by definition of altlook, that altlook(m, TA) = fail. By definition of mtype, we have that mtype(m, TA) is inferred from msig(m, TA) (rule (MTyp·Virt)), and that mtype(m, TA 1 ) is inferred from altlook(m 1 ) = I ? I, 1 ) (rule (MTyp·Impl)). By rule (MSig·Ex 1 ), we have mtype(m