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Proof step 10 from proof getbar-putbar-same-01 (previous step)

0 < 1 + n→i(# bytes0), ∀ r, i, st. getbytearray(r, i, n→i(# bytes0), putByteArray(bytes0, r, i, st)) = bytes0 ⊦ (putByteArray(by ' + bytes0, r, i, st)[r -- i ']) .byteval + getbytearray(r, i + 1, 1 + n→i(# bytes0) - 1, putByteArray(by ' + bytes0, r, i, st)) = by ' + bytes0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ r, i, st. getbytearray(r, i, n→i(# bytes0), putByteArray(bytes0, r, i, st)) = bytes0 ⊦ (putByteArray(by ' + bytes0, r, i, st)[r -- i ']) .byteval ' + getbytearray(r, i + 1, n→i(# bytes0), putByteArray(by ' + bytes0, r, i, st)) = by ' + bytes0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 ≤ n→i(n)], [CSIMPSEQ : ⊦ (~1) < i ↔ 0 ≤ i], [CSIMPSEQ : ⊦ ((* j)) < i0 + ((* j0)) ↔ j - j0 < i0], [CSIMPSEQ : ⊦ by + bytes = by ' + bytes], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + 0 = i], [CSIMPSEQ : ⊦ ((* i)) + i0 - ((* j)) = (i - j) + i0])]]