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Proof step 3 from proof put-rkvs-list (conclusion)

⊦ (st[rk, val][rk0, val0]) = st[(rk × val) ' + (rk0 × val0) ']

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ st[(rk × val) ' + rkvs] = (st[rk, val][rkvs])
Proof step 2:
⊦ st[(rk × val) '] = st[rk, val]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (rkvs + rkvs0) + rkvs1 = rkvs + rkvs0 + rkvs1])]]