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Proof step 11 from proof get-sublist (previous step)

j = i1 + 1, 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) ⊦ st[r -- i '] ' + sublist(0, m, getarray(r, i + 1, i1, st)) = getarray(r, i, n→i(m) + 1, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
n→i(m) ≤ i1, 0 ≤ i1, ∀ j0, r0, i0, st0. ¬ j0 < n→i(m) → sublist(0, m, getarray(r0, i0, j0, st0)) = getarray(r0, i0, n→i(m), st0) ⊦ st[r -- i '] ' + sublist(0, m, getarray(r, i + 1, i1, st)) = getarray(r, i, n→i(m) + 1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ (~1) < i ↔ 0 ≤ i], [CSIMPSEQ : ⊦ ((* j)) < i0 + ((* j0)) ↔ j - j0 < i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i < j + 1 ↔ i ≤ j])]]