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Proof step 12 from proof getarray-union (previous step)

∀ 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) ⊦ getarray(r, i, j, st ∪ st0) = getarray(r, i, j, st0), r ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
0 < j ⊦ getarray(r, i, j, st) = st[r -- i '] ' + getarray(r, i + 1, j - 1, st)
Proof step 2:
⊦ j ≤ 0 → getarray(r, i, j, st) = []
Proof step 11:
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

Rule argument

[crewrite-arg: getarray-offset-02 , F , [st, j, i, r] → [st ∪ st0, j, i, r] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i])], [EXTRATERMS : (st ∪ st0, j, i, r)]]