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Proof step 5 from proof put-rkvs-append (previous step)

∀ st, rkvs0. st[rkvs1 + rkvs0] = (st[rkvs1][rkvs0]) ⊦ (st[rkv .1, rkv .2][rkvs1 + rkvs0]) = (st[rkv .1, rkv .2][rkvs1][rkvs0])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
rkv = rk × val, ∀ st, rkvs0. st[rkvs1 + rkvs0] = (st[rkvs1][rkvs0]) ⊦ (st[rk, val][rkvs1 + rkvs0]) = (st[rk, val][rkvs1][rkvs0])

Rule argument

[speclemmaarg: ,rk = rkv .1 ∧ val = rkv .2 ↔ rkv = rk × val ,[rkv, rk, val] → [rkv, rk, val] ,T ,pair ,refkeyvalues ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ rk = rkv .1 ∧ val = rkv .2 ↔ rkv = rk × val])], [EXTRATERMS : (rkv, rk, val)]]