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Proof step 13 from proof bar-sublist (previous step)

0 + n ≤ # getbytearray(r, i + n→i(m), n→i(n), st), 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ # getbytearray(r, i, n→i(m), st) = m
Proof step 12:
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)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m ≤ m], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ 0 + m = m])]]