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Proof step 5 from proof lookup-put-rkvs-notin (previous step)

rkv = rk0 × val, ∀ st, rk. (st[rkvs0][rk]) = st[rk] ∨ rk ∈ rkvs0 ⊦ (st[rk0, val][rkvs0][rk]) = st[rk], rk ∈ rkv ', rk ∈ rkvs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ st, rk. (st[rkvs0][rk]) = st[rk] ∨ rk ∈ rkvs0 ⊦ (st[rk0, val][rkvs0][rk]) = st[rk], rk = rk0, rk ∈ rkvs0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ (rk0 × val) ' ↔ rk = rk0])]]