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Proof step 25 from proof bar-sublist (previous step)

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)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ 0 ≤ j0 ∧ j0 ≤ j → getbytearray(r, i, j, st) = getbytearray(r, i, j0, st) + getbytearray(r, i + j0, j - j0, st)
Proof step 24:
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)

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 ≤ n→i(n)], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i], [CSIMPFORWARD : ⊦ i ≤ j ∧ j < j0 → i < j0], [CSIMPSEQ : ⊦ i + j < j ↔ i < 0], [CSIMPSEQ : ⊦ ¬ n→i(n) < 0], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i])], [EXTRATERMS : (n→i(m), j, r, i, st)]]