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Proof step 7 from proof getarray-in-01 (previous step)

0 < i3, val ∈ getarray(r, i0, i3, st), ∀ st, i0, r, val, i. i→n(i) < i→n(i3 + 1) → (val ∈ getarray(r, i0, i, st) ∧ 0 < i → (∃ i2. ¬ i2 < i0 ∧ ¬ i0 + i ≤ i2 ∧ st[r -- i2 '] = val)) ⊦ ∃ i. ¬ i < i0 ∧ ¬ i0 + i3 < i ∧ st[r -- i '] = val

User applied rule apply induction

And obtained the following new goals

Proof step 6:
∃ i2. ¬ i2 < i0 ∧ ¬ i0 + i3 ≤ i2 ∧ st[r -- i2 '] = val, 0 < i3, val ∈ getarray(r, i0, i3, st), ∀ st, i0, r, val, i. i→n(i) < i→n(i3 + 1) → (val ∈ getarray(r, i0, i, st) ∧ 0 < i → (∃ i2. ¬ i2 < i0 ∧ ¬ i0 + i ≤ i2 ∧ st[r -- i2 '] = val)) ⊦ ∃ i. ¬ i < i0 ∧ ¬ i0 + i3 < i ∧ st[r -- i '] = val

Rule argument

[substlist-arg: [st, i0, r, val, i] → [st, i0, r, val, i3]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i < j → (¬ ¬ j ≤ i ↔ false)], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j)], [CSIMPSEQ : ⊦ 0 < j → abs(j) = j], [CSIMPSEQ : ⊦ ¬ i < i], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])], [EXTRATERMS : (st, i0, r, val, i3)]]