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Proof step 6 from proof at-1dim-01 (previous step)

i ≠ j + 1, 0 ≤ j, j < i, 0 < i, ∀ st, val, r, j, i0. i→n(i0) < i→n(i) → (j < i0 ∧ 0 ≤ j ∧ 0 < i0 → (addarray(r, val, i0, st)[r -- j ']) = val) ⊦ (addarray(r, val, i - 1, st)[r -- j ']) = val

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < i, j < i, 0 ≤ j, ∀ st, val, r, j, i0. i→n(i0) < i→n(i) → (j < i0 ∧ 0 ≤ j ∧ 0 < i0 → (addarray(r, val, i0, st)[r -- j ']) = val) ⊦ i = j + 1, (addarray(r, val, i - 1, st)[r -- j ']) = val

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]