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Proof step 21 from proof get-sublist (previous step)

0 < j, n→i(n) + n→i(m0) < j, ∀ n0, j0, r0, i0, st0. ¬ j0 < n→i(m0) + n→i(n0) → sublist(m0, n0, getarray(r0, i0, j0, st0)) = getarray(r0, i0 + n→i(m0), n→i(n0), st0) ⊦ sublist(m0, n, getarray(r, i + 1, j - 1, st)) = getarray(r, (n→i(m0) + 1) + i, n→i(n), st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
j = i1 + 1, 0 < j, n→i(n) + n→i(m0) < j, ∀ n0, j0, r0, i0, st0. ¬ j0 < n→i(m0) + n→i(n0) → sublist(m0, n0, getarray(r0, i0, j0, st0)) = getarray(r0, i0 + n→i(m0), n→i(n0), st0) ⊦ sublist(m0, n, getarray(r, i + 1, i1, st)) = getarray(r, (n→i(m0) + 1) + i, n→i(n), st)

Rule argument

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

Additional information

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