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Proof step 13 from proof getarray-in-01 (previous step)

1 < i1, 0 < i1, ∀ st, i0, r, val, i. i→n(i) < i→n(i1) → (val ∈ getarray(r, i0, i, st) ∧ 0 < i → (∃ i2. ¬ i2 < i0 ∧ ¬ i0 + i ≤ i2 ∧ st[r -- i2 '] = val)) ⊦ ∃ i. ¬ i < i0 ∧ ¬ i0 + i1 ≤ i ∧ st[r -- i '] = val, ¬ val ∈ getarray(r, i0, i1 - 1, st) ∧ val ≠ st[r -- (i0 + i1 - 1) ']

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
i1 = i3 + 1, 1 < i1, 0 < i1, ∀ st, i0, r, val, i. i→n(i) < i→n(i1) → (val ∈ getarray(r, i0, i, st) ∧ 0 < i → (∃ i2. ¬ i2 < i0 ∧ ¬ i0 + i ≤ i2 ∧ st[r -- i2 '] = val)) ⊦ ∃ i. ¬ i < i0 ∧ ¬ i0 + i1 ≤ i ∧ st[r -- i '] = val, ¬ val ∈ getarray(r, i0, i3, st) ∧ val ≠ st[r -- (i0 + i1 - 1) ']

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [i1, 1, i3] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (i1, 1, i3)]]