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Proof step 3 from proof sublist-getbar (previous step)

⊦ n→i(n) + n→i(n0) ≤ st[r -- _length] .intval → sublist(n, n0, getbytearray(r, 0, st[r -- _length] .intval, st)) = getbytearray(r, n→i(n), n→i(n0), st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ n→i(m) + n→i(n) ≤ j → sublist(m, n, getbytearray(r, i, j, st)) = getbytearray(r, i + n→i(m), n→i(n), st)

Rule argument

[crewrite-arg: bar-sublist , F , [st, j, i, r, n, m] → [st, st[r -- _length] .intval, 0, r, n0, n] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + 0 = i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i])], [EXTRATERMS : (st, st[r -- _length] .intval, 0, r, n0, n)]]