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Proof step 3 from proof in-1dim-15 (previous step)

i = i1 + 1, 0 ≤ j, j < i, 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)) ⊦ i = j + 1, r -- j ' ∈ addarray(r, val, i1, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
0 ≤ j, j < i1, 0 < i1, 0 ≤ i1, ∀ st, val, r, j, i0. i→n(i0) < i→n(i1 + 1) → (j < i0 ∧ 0 ≤ j ∧ 0 < i0 → r -- j ' ∈ addarray(r, val, i0, st)) ⊦ i1 = j, r -- j ' ∈ addarray(r, val, i1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ (~1) < i ↔ 0 ≤ i], [CSIMPSEQ : ⊦ ((* j)) < i0 + ((* j0)) ↔ j - j0 < i0], [CSIMPSEQ : ⊦ i ≠ j → (i ≤ j ↔ i < j)], [CSIMPSEQ : ⊦ i < j + 1 ↔ i ≤ j], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = i + i0 ↔ j = i0])]]