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Proof step 5 from proof get-length (previous step)

1 < j, 0 < j, ∀ j0, r, i, st. i→n(j0) < i→n(j) → (0 < j0 → # getarray(r, i, j0, st) = i→n(j0)) ⊦ # getarray(r, i, j - 1, st) + 1 = i→n(j)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
j = i0 + 1, 1 < j, 0 < j, ∀ j0, r, i, st. i→n(j0) < i→n(j) → (0 < j0 → # getarray(r, i, j0, st) = i→n(j0)) ⊦ # getarray(r, i, i0, st) + 1 = i→n(j)

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)]]