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Proof step 18 from proof bar-sublist (previous step)

0 ≤ i0, n→i(n) ≤ i0 ⊦ sublist(0, n, getbytearray(r, i + n→i(m), i0, st)) = getbytearray(r, i + n→i(m), n→i(n), st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 0 ≤ j0 ∧ j0 ≤ j → getbytearray(r, i, j, st) = getbytearray(r, i, j0, st) + getbytearray(r, i + j0, j - j0, st)
Proof step 17:
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)

Rule argument

[crewrite-arg: getbar-anyrec , F , [j0, j, r, i, st] → [n→i(n), i0, r, i + n→i(m), st] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 ≤ n→i(n)], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i], [CSIMPSEQ : ⊦ i ≤ j → (¬ ¬ j < i ↔ false)], [CSIMPSEQ : ⊦ i + j = j + i])], [EXTRATERMS : (n→i(n), i0, r, i + n→i(m), st)]]