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Proof step 11 from proof getbar-restn-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
st[r -- _length] .intval = i + n→i(n), 0 ≤ i ⊦ restn(n, getbytearray(r, 0, n→i(n), st) + getbytearray(r, n→i(n), i, st)) = getbytearray(r, n→i(n), i, st)

Rule argument

[emptyarg]

Additional information

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