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Proof step 15 from proof at=get (previous step)

1 < j, true, ∀ r, st, i, j0. i→n(j0) < i→n(j) → (true → (0 ≤ i ∧ i < j0 → st[r -- i '] = (getarray(r, 0, j0, st)[i→n(i)]))) ⊦ 0 ≤ i ∧ i < j → st[r -- i '] = (getarray(r, 0, j - 1, st) + st[r -- (0 + j - 1) '])[i→n(i)]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
i < j, 0 ≤ i, 1 < j, 0 < j, ∀ r, st, i, j0. i→n(j0) < i→n(j) → (true → (0 ≤ i ∧ i < j0 → st[r -- i '] = (getarray(r, 0, j0, st)[i→n(i)]))) ⊦ st[r -- i '] = (getarray(r, 0, j - 1, st) + st[r -- (j - 1) '] ')[i→n(i)]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPFORWARD : ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j], [CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPSEQ : ⊦ vals + val = vals + val '], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + 0 = i])]]