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Proof step 23 from proof bar-sublist (previous step)

n→i(m) ≤ j, n→i(m) + n→i(n) ≤ j ⊦ sublist(m, n, getbytearray(r, i, n→i(m), st) + getbytearray(r, i + n→i(m), j - n→i(m), 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 22:
j = i0 + n→i(m), n→i(m) ≤ j, n→i(m) + n→i(n) ≤ j ⊦ sublist(m, n, getbytearray(r, i, n→i(m), st) + getbytearray(r, i + n→i(m), i0, st)) = getbytearray(r, i + n→i(m), n→i(n), st)

Rule argument

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

Additional information

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