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Proof step 8 from proof in-1dim-02 (previous step)

true, ∀ 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, st) ↔ r -- sk ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ 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, st) ↔ r -- sk ∈ st, r = r0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]