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Proof step 15 from proof bar-sublist (previous step)

i0 = i1 + n→i(n), 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), i1, st)) = getbytearray(r, i + n→i(m), n→i(n), st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
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), i1, st)) = getbytearray(r, i + n→i(m), n→i(n), st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i ≤ i + j ↔ 0 ≤ j])]]