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Proof step 8 from proof at=get (previous step)

¬ i→n(i) < # getarray(r, 0, i0, st), i ≤ i0, 0 ≤ i, 0 ≤ i0, 0 < i0, ∀ r, st, i, j0. i→n(j0) < i→n(i0 + 1) → (true → (0 ≤ i ∧ i < j0 → st[r -- i '] = (getarray(r, 0, j0, st)[i→n(i)]))) ⊦ st[r -- i '] = (getarray(r, 0, i0, st) + st[r -- i0 '] ')[i→n(i)]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ 0 < j → # getarray(r, i, j, st) = i→n(j)
Proof step 7:
0 < i0, 0 ≤ i0, ∀ r, st, i, j0. i→n(j0) < i→n(i0 + 1) → (true → (0 ≤ i ∧ i < j0 → st[r -- i '] = (getarray(r, 0, j0, st)[i→n(i)]))) ⊦ st[r -- i0 '] = (getarray(r, 0, i0, st) + st[r -- i0 '] ')[i→n(i0)]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i ≤ i], [CSIMPSEQ : ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j], [CSIMPSEQ : ⊦ i ≤ j → (j ≤ i ↔ i = j)], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ 0 < j → abs(j) = j], [CSIMPSEQ : ⊦ 0 ≤ i → abs(i) = i], [CSIMPSEQ : ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j)])]]