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Proof step 4 from proof bar-restn (previous step)

n→i(m) < j, 0 ≤ i, 0 < j, ∀ j0, i0, r0, st0. ¬ j0 < n→i(m) ∧ ¬ i0 < 0 → restn(m, bytevalues(getarray(r0, i0, j0, st0))) = bytevalues(getarray(r0, i0 + n→i(m), j0 - n→i(m), st0)) ⊦ restn(m, bytevalues(getarray(r, i + 1, j - 1, st))) = bytevalues(getarray(r, (n→i(m) + 1) + i, j - n→i(m) + 1, st))

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 4, [QUANTTERMLIST : (j - 1, i + 1, r, st)]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i - j < i0 ↔ i < j + i0], [CSIMPSEQ : ⊦ i < j + 1 ↔ i ≤ j], [CSIMPSEQ : ⊦ i < j → (¬ ¬ j ≤ i ↔ false)], [CSIMPSEQ : ⊦ 0 ≤ i ∧ 0 ≤ j → ¬ i + j < 0], [CSIMPSEQ : ⊦ i - j - i0 = i - j + i0])], [EXTRATERMS : (j - 1, i + 1, r, st)]]