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Proof step 15 from proof putbar-getbar-le (previous step)

i→n(j) = m + 1 ∧ i→n(j) ≠ 0, 0 < j, j ≤ 1 + n→i(# bytes0), ∀ j0, r0, i0, st0. ¬ j0 < 0 ∧ ¬ n→i(# bytes0) < j0 → getbytearray(r0, i0, j0, putByteArray(bytes0, r0, i0, st0)) = firstn(i→n(j0), bytes0) ⊦ j = 0, getbytearray(r, i, j, putByteArray(bytes0, r, i + 1, st)[r -- i ', byteval(by)]) = by ' + firstn(m, bytes0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i→n(j) = m + 1, 0 < j, j ≤ 1 + n→i(# bytes0), ∀ j0, r0, i0, st0. ¬ j0 < 0 ∧ ¬ n→i(# bytes0) < j0 → getbytearray(r0, i0, j0, putByteArray(bytes0, r0, i0, st0)) = firstn(i→n(j0), bytes0) ⊦ j = 0, getbytearray(r, i, j, putByteArray(bytes0, r, i + 1, st)[r -- i ', byteval(by)]) = by ' + firstn(m, bytes0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i→n(i) = 0 ↔ i = 0])]]