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Proof step 7 from proof in-put-rkvs (previous step)

¬ ¬ rk0 ∈ rkvs0, ∀ rk, st. rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk ∈ rkvs0 ⊦ rk ∈ (st[rk0, val][rkvs0]) ↔ rk ∈ st ∨ rk = rk0 ∨ rk ∈ rkvs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ rk ∈ rkvs → (st[rk, val][rkvs]) = st[rkvs]
Proof step 6:
rk0 ∈ rkvs0, ∀ rk, st. rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk ∈ rkvs0 ⊦ rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk ∈ rkvs0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]