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Proof step 24 from proof bar-sublist (previous step)

0 ≤ 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), j - n→i(m), st)) = getbytearray(r, i + n→i(m), n→i(n), st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
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)

Rule argument

[emptyarg]

Additional information

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