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Proof step 18 from proof putArray-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
∀ 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

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]