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Proof step 9 from proof get-first (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) .first = st[r -- i ']) ⊦ (getarray(r, i, i0, st) + st[r -- (i + j - 1) '] ') .first = st[r -- i ']

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 < i0, 0 ≤ i0, ∀ j0, r, i, st. i→n(j0) < i→n(i0 + 1) → (0 < j0 → getarray(r, i, j0, st) .first = st[r -- i ']) ⊦ (getarray(r, i, i0, st) + st[r -- (i + i0) '] ') .first = st[r -- i ']

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 1 ≤ i ↔ 0 < i], [CSIMPSEQ : ⊦ i < j + 1 ↔ i ≤ j], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j - i = j])]]