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Proof step 6 from proof putar-in-02 (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 5:
r -- j ' ∈ putArray(vals0, r, i + 1, st) ↔ r -- j ' ∈ st ∨ i + 1 ≤ j ∧ j < (i + 1) + n→i(# vals0), ∀ r0, j0, i0, st0. r0 -- j0 ' ∈ putArray(vals0, r0, i0, st0) ↔ r0 -- j0 ' ∈ st0 ∨ i0 ≤ j0 ∧ j0 < i0 + n→i(# vals0) ⊦ r -- j ' ∈ putArray(vals0, r, i + 1, st) ↔ r -- j ' ∈ st ∨ ¬ n→i(# vals0) + i < j ∧ ¬ j < i, j = i

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (r, j, i + 1, st) abort:F , discard:F , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, j, i + 1, st)]]