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Proof step 6 from proof getbar-restn-04 (previous step)

n ≤ # getbytearray(r, i, i0 + n→i(n), st), 0 ≤ i, 0 ≤ i0 + n→i(n), 0 ≤ i0 ⊦ mkjavavalues(restn(n, getbytearray(r, i, i0 + n→i(n), st))) = mkjavavalues(getbytearray(r, n→i(n) + i, i0, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ 0 ≤ j → # getbytearray(r, i, j, st) = i→n(j)
Proof step 5:
0 ≤ i, 0 ≤ i0 + n→i(n), 0 ≤ i0 ⊦ mkjavavalues(restn(n, getbytearray(r, i, i0 + n→i(n), st))) = mkjavavalues(getbytearray(r, n→i(n) + i, i0, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m ≤ m + n], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ 0 ≤ i → i→n(i + n→i(n)) = i→n(i) + n])]]