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Proof step 4 from proof lookup-1dim-01 (previous step)

i = i1 + 1, 0 < i, ∀ sk, st, val, r0, r, i0. i→n(i0) < i→n(i) → (true → r = r0 ∨ (addarray(r, val, i0, st)[r0 -- sk]) = st[r0 -- sk]) ⊦ r = r0, (addarray(r, val, i1, st)[r0 -- sk]) = st[r0 -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 ≤ i1, ∀ sk, st, val, r0, r, i0. i→n(i0) < i→n(i1 + 1) → (true → r = r0 ∨ (addarray(r, val, i0, st)[r0 -- sk]) = st[r0 -- sk]) ⊦ r = r0, (addarray(r, val, i1, st)[r0 -- sk]) = st[r0 -- sk]

Rule argument

[emptyarg]

Additional information

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