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Proof step 10 from proof add-put-switch (previous step)

is_fskey(sk), 0 < i, ∀ val0, st, val, r0, sk, i0. i→n(i0) < i→n(i) → (is_fskey(sk) → addarray(r0, val, i0, st[r0 -- sk, val0]) = (addarray(r0, val, i0, st)[r0 -- sk, val0])) ⊦ (addarray(r0, val, i - 1, st[r0 -- sk, val0])[r0 -- (i - 1) ', val]) = (addarray(r0, val, i, st)[r0 -- sk, val0])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
i = i1 + 1, is_fskey(sk), 0 < i, ∀ val0, st, val, r0, sk, i0. i→n(i0) < i→n(i) → (is_fskey(sk) → addarray(r0, val, i0, st[r0 -- sk, val0]) = (addarray(r0, val, i0, st)[r0 -- sk, val0])) ⊦ (addarray(r0, val, i1, st[r0 -- sk, val0])[r0 -- i1 ', val]) = (addarray(r0, val, i, st)[r0 -- sk, val0])

Rule argument

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

Additional information

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