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Proof step 15 from proof getintarray-in-02 (previous step)

0 ≤ i1, ∀ st, i0, r, i, j0. i→n(j0) < i→n(i1 + 1) → (true → (i ∈ getintarray(r, i0, j0, st) ↔ (∃ j1. i = st[r -- (i0 + j1) '] .intval ∧ ¬ j1 < 0 ∧ ¬ j0 ≤ j1))) ⊦ i ∈ getintarray(r, i0, i1, st) ∨ i = st[r -- (i0 + i1) '] .intval ↔ (∃ j. i = st[r -- (i0 + j) '] .intval ∧ ¬ j < 0 ∧ ¬ i1 < j)

User applied rule apply induction

And obtained the following new goals

Proof step 14:
i ∈ getintarray(r, i0, i1, st) ↔ (∃ j1. i = st[r -- (i0 + j1) '] .intval ∧ ¬ j1 < 0 ∧ ¬ i1 ≤ j1), 0 ≤ i1, ∀ st, i0, r, i, j0. i→n(j0) < i→n(i1 + 1) → (true → (i ∈ getintarray(r, i0, j0, st) ↔ (∃ j1. i = st[r -- (i0 + j1) '] .intval ∧ ¬ j1 < 0 ∧ ¬ j0 ≤ j1))) ⊦ i ∈ getintarray(r, i0, i1, st) ∨ i = st[r -- (i0 + i1) '] .intval ↔ (∃ j. i = st[r -- (i0 + j) '] .intval ∧ ¬ j < 0 ∧ ¬ i1 < j)

Rule argument

[substlist-arg: [st, i0, r, i, j0] → [st, i0, r, i, i1]]

Additional information

[[CONCRETESIMPRULES : ([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 ≤ i → abs(i) = i], [CSIMPSEQ : ⊦ ¬ i < i], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])], [EXTRATERMS : (st, i0, r, i, i1)]]