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

∀ r1, r0, i0, st. r1 ∈ putArray(vals0, r0, i0, st) ↔ r1 ∈ st ∨ r1 = r0 ∧ vals0 ≠ [] ⊦ r1 ∈ (putArray(vals0, r0, i0 + 1, st)[r0 -- i0 ', val]) ↔ r1 = r0 ∨ r1 ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ r ∈ st[r0 -- sk, val] ↔ r = r0 ∨ r ∈ st
Proof step 5:
∀ r1, r0, i0, st. r1 ∈ putArray(vals0, r0, i0, st) ↔ r1 ∈ st ∨ r1 = r0 ∧ vals0 ≠ [] ⊦ r1 = r0 ∨ r1 ∈ putArray(vals0, r0, i0 + 1, st) ↔ r1 = r0 ∨ r1 ∈ st

Rule argument

[speclemmaarg: ,r ∈ st[r0 -- sk, val] ↔ r = r0 ∨ r ∈ st ,[val, sk, r0, st, r] → [val, i0 ', r0, putArray(vals0, r0, i0 + 1, st), r1] ,F ,storefuns , ,instore-02 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val, i0 ', r0, putArray(vals0, r0, i0 + 1, st), r1)]]