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Proof step 7 from proof in-1dim-15 (previous step)

j < i ∧ 0 ≤ j ∧ 0 < i, ∀ st, val, r, j, i0. i→n(i0) < i→n(i) → (j < i0 ∧ 0 ≤ j ∧ 0 < i0 → r -- j ' ∈ addarray(r, val, i0, st)) ⊦ r -- j ' ∈ addarray(r, val, i, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < i, 0 ≤ j, j < i, ∀ st, val, r, j, i0. i→n(i0) < i→n(i) → (j < i0 ∧ 0 ≤ j ∧ 0 < i0 → r -- j ' ∈ addarray(r, val, i0, st)) ⊦ r -- j ' ∈ addarray(r, val, i, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]