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Proof step 17 from proof getintarray-in-02 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
j = i1 + 1, 0 < j, ∀ st, i0, r, i, j0. i→n(j0) < i→n(j) → (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 + j - 1) '] .intval ↔ (∃ j0. i = st[r -- (i0 + j0) '] .intval ∧ ¬ j0 < 0 ∧ ¬ j ≤ j0)

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [j, 1, i1] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (j, 1, i1)]]