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Proof step 15 from proof get-sublist (previous step)

∀ j, r, i, st. n→i(m) ≤ j → sublist(0, m, getarray(r, i, j, st)) = getarray(r, i, n→i(m), st) ⊦ n→i(m +1) ≤ j → sublist(0, m +1, getarray(r, i, j, st)) = getarray(r, i, n→i(m +1), st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
n→i(m) < j, 0 < j, ∀ j0, r0, i0, st0. ¬ j0 < n→i(m) → sublist(0, m, getarray(r0, i0, j0, st0)) = getarray(r0, i0, n→i(m), st0) ⊦ sublist(0, m + 1, getarray(r, i, j, st)) = getarray(r, i, n→i(m) + 1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ n→i(m) < j → 0 < j], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + 1 ≤ j ↔ i < j], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ n→i(m + n) = n→i(m) + n→i(n)], [CSIMPSEQ : ⊦ m +1 = m + 1])]]