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Proof step 13 from proof put-rkvs-absorb (previous step)

rk ∈ rkvs1 + (rk0 × val0) ', ∀ val, st, rk, rkvs0. # rkvs0 < # (rkvs1 + (rk0 × val0) ') → (rk ∈ rkvs0 → (st[rk, val][rkvs0]) = st[rkvs0]) ⊦ (st[rk, val][rkvs1][rk0, val0]) = (st[rkvs1][rk0, val0])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ rk ∈ rkvs + rkvs0 ↔ rk ∈ rkvs ∨ rk ∈ rkvs0
Proof step 12:
rk ∈ rkvs1 ∨ rk ∈ (rk0 × val0) ', ∀ val, st, rk, rkvs0. # rkvs0 < # (rkvs1 + (rk0 × val0) ') → (rk ∈ rkvs0 → (st[rk, val][rkvs0]) = st[rkvs0]) ⊦ (st[rk, val][rkvs1][rk0, val0]) = (st[rkvs1][rk0, val0])

Rule argument

[speclemmaarg: ,rk ∈ rkvs + rkvs0 ↔ rk ∈ rkvs ∨ rk ∈ rkvs0 ,[rkvs0, rkvs, rk] → [(rk0 × val0) ', rkvs1, rk] ,F ,assoclist ,refkeyvalues ,in-append ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : ((rk0 × val0) ', rkvs1, rk)]]