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Proof step 9 from proof in-1dim-13 (previous step)

r0 -- sk ∈ st ∧ r0 ∈ st, ∀ val, r, st, sk, r0, i0. i→n(i0) < i→n(i) → (r0 -- sk ∈ st ∧ r0 ∈ st → r0 -- sk ∈ addarray(r, val, i0, st)) ⊦ r0 -- sk ∈ addarray(r, val, i, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
r0 ∈ st, r0 -- sk ∈ st, ∀ val, r, st, sk, r0, i0. i→n(i0) < i→n(i) → (r0 -- sk ∈ st ∧ r0 ∈ st → r0 -- sk ∈ addarray(r, val, i0, st)) ⊦ r0 -- sk ∈ addarray(r, val, i, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]