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Proof step 15 from proof putArray-in (previous step)

∀ rk, r, i, st. rk ∈ putArray(vals0, r, i, st) ↔ rk ∈ st ∨ rk .ref = r ∧ is_indexkey(rk .key) ∧ i ≤ rk .key .index ∧ rk .key .index < i + n→i(# vals0) ⊦ r0 -- sk ∈ putArray(vals0, r, i + 1, st[r -- i ', val]) ↔ r0 = r ∧ ¬ is_fskey(sk) ∧ ¬ sk .index < i ∧ ¬ n→i(# vals0) + i < sk .index ∨ r0 -- sk ∈ st

User applied rule all left

And obtained the following new goals

Proof step 14:
r0 -- sk ∈ putArray(vals0, r, i + 1, st[r -- i ', val]) ↔ r0 -- sk ∈ st[r -- i ', val] ∨ (r0 -- sk) .ref = r ∧ is_indexkey((r0 -- sk) .key) ∧ i + 1 ≤ (r0 -- sk) .key .index ∧ (r0 -- sk) .key .index < (i + 1) + n→i(# vals0), ∀ rk, r, i, st. rk ∈ putArray(vals0, r, i, st) ↔ rk ∈ st ∨ rk .ref = r ∧ is_indexkey(rk .key) ∧ i ≤ rk .key .index ∧ rk .key .index < i + n→i(# vals0) ⊦ r0 -- sk ∈ putArray(vals0, r, i + 1, st[r -- i ', val]) ↔ r0 = r ∧ ¬ is_fskey(sk) ∧ ¬ sk .index < i ∧ ¬ n→i(# vals0) + i < sk .index ∨ r0 -- sk ∈ st

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (r0 -- sk, r, i + 1, st[r -- i ', val]) abort:F , discard:F , computed:T ]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r0 -- sk, r, i + 1, st[r -- i ', val])]]