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Proof step 4 from proof get-length (previous step)

j = i0 + 1, 1 < j, 0 < j, ∀ j0, r, i, st. i→n(j0) < i→n(j) → (0 < j0 → # getarray(r, i, j0, st) = i→n(j0)) ⊦ # getarray(r, i, i0, st) + 1 = i→n(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
0 < i0, ∀ j0, r, i, st. i→n(j0) < i→n(i0 + 1) → (0 < j0 → # getarray(r, i, j0, st) = i→n(j0)) ⊦ # getarray(r, i, i0, st) = i→n(i0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPSEQ : ⊦ 0 < i ∧ 0 < j → 0 < i + j], [CSIMPSEQ : ⊦ ((* j)) < i0 + ((* j0)) ↔ j - j0 < i0], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0], [CSIMPSEQ : ⊦ 0 < i + 1 → i→n(i + 1) = i→n(i) + 1], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0])]]