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Proof step 11 from proof shift-1dim-01 (previous step)

∀ val, sk, st, val0, r0, r, i0. i→n(i0) < i→n(i) → (true → r = r0 ∨ (addarray(r, val0, i0, st)[r0 -- sk, val]) = addarray(r, val0, i0, st[r0 -- sk, val])) ⊦ (addarray(r, val0, i, st)[r0 -- sk, val]) = addarray(r, val0, i, st[r0 -- sk, val]), r = r0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ i > 0 → addarray(r, val, i, st) = (addarray(r, val, i - 1, st)[r -- (i - 1) ', val])
Proof step 2:
⊦ i ≤ 0 → addarray(r, val, i, st) = st
Proof step 10:
i > 0, ∀ val, sk, st, val0, r0, r, i0. i→n(i0) < i→n(i) → (true → r = r0 ∨ (addarray(r, val0, i0, st)[r0 -- sk, val]) = addarray(r, val0, i0, st[r0 -- sk, val])) ⊦ (addarray(r, val0, i - 1, st)[r -- (i - 1) ', val0][r0 -- sk, val]) = addarray(r, val0, i, st[r0 -- sk, val]), r = r0

Rule argument

[crewrite-arg: addarray-rec, F, [st, i, val, r] → [st, i, val0, r]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i > j ↔ j < i], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i])], [EXTRATERMS : (st, i, val0, r)]]