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Proof step 10 from proof get-sublist (previous step)

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)

User applied rule all left

And obtained the following new goals

Proof step 9:
n→i(m) ≤ i1, 0 ≤ i1, ¬ i1 < n→i(m) → sublist(0, m, getarray(r, i + 1, i1, st)) = getarray(r, i + 1, n→i(m), st) ⊦ st[r -- i '] ' + sublist(0, m, getarray(r, i + 1, i1, st)) = getarray(r, i, n→i(m) + 1, st)

Rule argument

[EXRARG : pos: left 3, [extquanttermlist: (i1, r, i + 1, st) abort:F , discard:T , computed:F ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (i1, r, i + 1, st)]]