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Proof step 7 from proof lookup-put-rkvs-notin (previous step)

∀ st, rk. (st[rkvs0][rk]) = st[rk] ∨ rk ∈ rkvs0 ⊦ (st[rkv + rkvs0][rk]) = st[rk], rk ∈ rkv + rkvs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ st[rkvs + rkvs0] = (st[rkvs][rkvs0])
Proof step 3:
⊦ st[rkv '] = st[rkv .1, rkv .2]
Proof step 6:
∀ st, rk. (st[rkvs0][rk]) = st[rk] ∨ rk ∈ rkvs0 ⊦ (st[rkv .1, rkv .2][rkvs0][rk]) = st[rk], rk ∈ rkv ', rk ∈ rkvs0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ rkvs + rkvs0 ↔ rk ∈ rkvs ∨ rk ∈ rkvs0], [CSIMPSEQ : ⊦ (rkvs + rkvs0) + rkvs1 = rkvs + rkvs0 + rkvs1], [CSIMPSEQ : ⊦ rkv + rkvs = rkv ' + rkvs])]]