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Proof step 6 from proof copy-bet-low (previous step)

i0 + j = j0 + 1, 0 < j, i ≤ i0, i0 ≤ j0, i ≤ j0, ∀ j1, i, i0, j0, r, st. i→n(j1) < i→n(j) → (true → (i ≤ i0 ∧ i0 ≤ j0 ∧ j0 < i0 + j1 → (arraycopy(r, i, r, i0, j1, st)[r -- j0 ']) = st[r -- ((j0 - i0) + i) '])) ⊦ (arraycopy(r, i, r, i0, j - 1, st[r -- j0 ', st[r -- (i + j - 1) ']])[r -- j0 ']) = st[r -- (i + j0 - i0) ']

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ j = i0 + j0 ↔ j - j0 = i0
Proof step 5:
i0 + j - 1 = j0, 0 < j, i ≤ i0, i0 ≤ j0, i ≤ j0, ∀ j1, i, i0, j0, r, st. i→n(j1) < i→n(j) → (true → (i ≤ i0 ∧ i0 ≤ j0 ∧ j0 < i0 + j1 → (arraycopy(r, i, r, i0, j1, st)[r -- j0 ']) = st[r -- ((j0 - i0) + i) '])) ⊦ (arraycopy(r, i, r, i0, j - 1, st[r -- j0 ', st[r -- (i + j - 1) ']])[r -- j0 ']) = st[r -- (i + j0 - i0) ']

Rule argument

[speclemmaarg: ,((* j)) = i0 + ((* j0)) ↔ j - j0 = i0 ,[j0, i0, j] → [1, j0, i0 + j] ,F ,int-abs , ,numadd-01 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (1, j0, i0 + j)]]