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Proof step 6 from proof in-1dim-02 (previous step)

i > 0, ∀ 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 -- sk ∈ (addarray(r0, val, i - 1, st)[r0 -- (i - 1) ', val]) ↔ r -- sk ∈ st, r = r0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
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, i - 1, st) ↔ r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPSEQ : ⊦ i > j ↔ j < i], [CSIMPSEQ : ⊦ r ≠ r0 → ( r -- sk ∈ st[r0 -- sk0, val] ↔ r -- sk ∈ st)])]]