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Proof step 16 from proof bar-sublist (previous step)

0 ≤ i0, n→i(n) ≤ i0 ⊦ sublist(0, n, getbytearray(r, i + n→i(m), n→i(n), st) + getbytearray(r, (i + n→i(m)) + n→i(n), i0 - n→i(n), st)) = getbytearray(r, i + n→i(m), n→i(n), st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
i0 = i1 + n→i(n), 0 ≤ i0, n→i(n) ≤ i0 ⊦ sublist(0, n, getbytearray(r, i + n→i(m), n→i(n), st) + getbytearray(r, (i + n→i(m)) + n→i(n), i1, st)) = getbytearray(r, i + n→i(m), n→i(n), st)

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [i0, n→i(n), i1] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (i0, n→i(n), i1)]]