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Proof step 10 from proof in-put-rkvs (previous step)

∀ rk, st. rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk ∈ rkvs0 ⊦ rk = rk0 ∨ rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk = rk0 ∨ rk ∈ rkvs0, rk0 ∈ rkvs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ rk, st. rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk ∈ rkvs0 ⊦ rk ∈ st[rkvs0] ↔ rk ∈ st ∨ rk ∈ rkvs0, rk = rk0, rk0 ∈ rkvs0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]