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Proof step 12 from proof put-rkvs-absorb (previous step)

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])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
rk ∈ rkvs1 ∨ rk = rk0, ∀ 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

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ (rk0 × val) ' ↔ rk = rk0])]]