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Proof step 14 from proof put-rkvs-absorb (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ st[rkvs + rkvs0] = (st[rkvs][rkvs0])
Proof step 4:
⊦ st[(rk × val) '] = st[rk, val]
Proof step 13:
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])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (rkvs + rkvs0) + rkvs1 = rkvs + rkvs0 + rkvs1], [CSIMPSEQ : ⊦ rkvs + rkv ' ≠ []])]]