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Proof step 5 from proof getarray-union-03 (previous step)

0 < j, 0 ≤ i, r ∈ st, j + i ≤ st[r -- _length] .intval, r -- _length ∈ st, ∀ i0. ¬ i0 < i ∧ ¬ i + j ≤ i0 → r -- i0 ' ∈ st, ∀ st0, i, st, r, j0. i→n(j0) < i→n(j) → ( true → ( r -- _length ∈ st ∧ 0 ≤ i ∧ j0 ≤ st[r -- _length] .intval - i ∧ (∀ i1. i ≤ i1 ∧ i1 < i + j0 → r -- i1 ' ∈ st) → getarray(r, i, j0, st0 ∪ st) = getarray(r, i, j0, st))) ⊦ (st0 ∪ st)[r -- i '] ' + getarray(r, i + 1, j - 1, st0 ∪ st) = st[r -- i '] ' + getarray(r, i + 1, j - 1, st), r -- i ' ∈ st .keys

User applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 6, [extquanttermlist: (i)abort:F, discard:T, computed:F]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ i < i], [CSIMPSEQ : ⊦ i + j ≤ i ↔ j ≤ 0], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i < j → (¬ ¬ j ≤ i ↔ false)], [CSIMPSEQ : ⊦ rk ∈ st0 → (st ∪ st0)[rk] = st0[rk]], [CSIMPSEQ : ⊦ vals + vals0 = vals + vals1 ↔ vals0 = vals1], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ rk ∈ st → (¬ rk ∈ st .keys ↔ false)])], [EXTRATERMS : (i)]]