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

∀ rk, rk0, st, val. rk = rk0 ∨ (st[rk, val][rkvs0][rk0]) = (st[rkvs0][rk0]) ⊦ rk = rk0, (st[rk, val][rkv .1, rkv .2][rkvs0][rk0]) = (st[rkv .1, rkv .2][rkvs0][rk0])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
rkv = rk1 × val0, ∀ rk, rk0, st, val. rk = rk0 ∨ (st[rk, val][rkvs0][rk0]) = (st[rkvs0][rk0]) ⊦ rk = rk0, (st[rk, val][rk1, val0][rkvs0][rk0]) = (st[rk1, val0][rkvs0][rk0])

Rule argument

[speclemmaarg: ,rk = rkv .1 ∧ val = rkv .2 ↔ rkv = rk × val ,[rkv, rk, val] → [rkv, rk1, val0] ,T ,pair ,refkeyvalues ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ rk = rkv .1 ∧ val = rkv .2 ↔ rkv = rk × val])], [EXTRATERMS : (rkv, rk1, val0)]]