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Proof step 4 from proof in-1dim-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

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])]]