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Proof step 12 from proof getbytearray-subset (previous step)

n→i(n) < j, n < i→n(j), 0 ≤ i, st ⊆ st0, 0 < j, ∀ j0. ¬ j0 < 0 ∧ ¬ i + j ≤ j0 → r -- j0 ' ∈ st ⊦ j = 0, noval .byteval = st0[r -- (i + n→i(n)) '] .byteval, r -- (i + n→i(n)) ' ∈ st

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 6, [QUANTTERMLIST : (i + n→i(n))]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j ≤ i0 + j ↔ i ≤ i0], [CSIMPSEQ : ⊦ i < j → (¬ ¬ j ≤ i ↔ false)], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPFORWARD : ⊦ i < i0 ∧ i0 < i1 → i < i1], [CSIMPFORWARD : ⊦ i < j ∧ j ≤ j0 → i < j0], [CSIMPFORWARD : ⊦ j < i ∧ i ≤ 0 → j < 0], [CSIMPSEQ : ⊦ i + j < j ↔ i < 0], [CSIMPSEQ : ⊦ ¬ n→i(n) < 0])], [EXTRATERMS : (i + n→i(n))]]