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Proof step 11 from proof getarray-union (previous step)

0 < j, ∀ st0, st, i, r, j0. i→n(j0) < i→n(j) → (true → getarray(r, i, j0, st ∪ st0) = getarray(r, i, j0, st0) ∨ r ∈ st) ⊦ (st ∪ st0)[r -- i '] ' + getarray(r, i + 1, j - 1, st ∪ st0) = getarray(r, i, j, st0), r ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
j = i0 + 1, 0 < j, ∀ st0, st, i, r, j0. i→n(j0) < i→n(j) → (true → getarray(r, i, j0, st ∪ st0) = getarray(r, i, j0, st0) ∨ r ∈ st) ⊦ (st ∪ st0)[r -- i '] ' + getarray(r, i + 1, i0, st ∪ st0) = getarray(r, i, j, st0), r ∈ st

Rule argument

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

Additional information

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