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Proof step 4 from proof putAr-get-other (previous step)

# vals0 + i→n(i) < i→n(j), 0 ≤ i, ∀ i0, j0, r0, st0. ¬ i→n(j0) < i→n(i0) + # vals0 ∧ ¬ i0 < 0 → (putArray(vals0, r0, i0, st0)[r0 -- j0 ']) = st0[r0 -- j0 '] ⊦ (putArray(vals0, r, i + 1, st)[r -- j ']) = st[r -- j '], j = 0, j = i

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ m < n → (n ≤ m ↔ false)], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ 0 ≤ i ∧ 0 ≤ j → ¬ i + j < 0])], [EXTRATERMS : (i + 1, j, r, st)]]