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Proof step 12 from proof bar-sublist (previous step)

0 ≤ i1 + n→i(n), 0 ≤ i1 ⊦ sublist(0, n, getbytearray(r, i + n→i(m), n→i(n), st)) = getbytearray(r, i + n→i(m), n→i(n), st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ # bytes = n → sublist(0, n, bytes) = bytes
Proof step 11:
⊦ # getbytearray(r, i, n→i(m), st) = m

Rule argument

[speclemmaarg: ,# bytes = n → sublist(0, n, bytes) = bytes ,[bytes, n] → [getbytearray(r, i + n→i(m), n→i(n), st), n] ,F ,list-del ,bytelist ,sublist-all ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j = j + i])], [EXTRATERMS : (getbytearray(r, i + n→i(m), n→i(n), st), n)]]