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Proof step 19 from proof get-sublist (previous step)

n→i(n) + n→i(m0) ≤ i1, 0 ≤ i1, ∀ 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)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 3, [QUANTTERMLIST : (n, i1, r, i + 1, st)]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i ≤ j → (¬ ¬ j < i ↔ false)])], [EXTRATERMS : (n, i1, r, i + 1, st)]]