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Proof step 17 from proof bar-sublist (previous step)

0 ≤ n→i(n) ∧ n→i(n) ≤ i0, 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 simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
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)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 ≤ n→i(n)])]]