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Proof step 18 from proof at=get-offset (previous step)

0 ≤ i0 ∧ 0 < j + i0 ∧ 0 < j ∧ i1 < j ∧ 0 ≤ i1 + i0 ∧ 0 ≤ i1, ∀ r, st, i1, i0, j0. i→n(j0) < i→n(j) → (0 ≤ i0 ∧ 0 < j0 + i0 ∧ 0 < j0 ∧ i1 < j0 ∧ 0 ≤ i1 + i0 ∧ 0 ≤ i1 → st[r -- (i1 + i0) '] = (getarray(r, i0, j0, st)[i→n(i1)])) ⊦ st[r -- (i1 + i0) '] = (getarray(r, i0, j, st)[i→n(i1)])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
0 ≤ i1, 0 ≤ i1 + i0, i1 < j, 0 < j, 0 < j + i0, 0 ≤ i0, ∀ r, st, i1, i0, j0. i→n(j0) < i→n(j) → (0 ≤ i0 ∧ 0 < j0 + i0 ∧ 0 < j0 ∧ i1 < j0 ∧ 0 ≤ i1 + i0 ∧ 0 ≤ i1 → st[r -- (i1 + i0) '] = (getarray(r, i0, j0, st)[i→n(i1)])) ⊦ st[r -- (i1 + i0) '] = (getarray(r, i0, j, st)[i→n(i1)])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]