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Proof step 7 from proof put-rkvs-switch (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ val, rk, st, rkvs0. # rkvs0 < # (rkvs1 + (rk0 × val0) ') → (true → (st[rk, val][rkvs0]) = (st[rkvs0][rk, val]) ∨ rk ∈ rkvs0) ⊦ rk = rk0, (st[rk, val][rkvs1][rk0, val0]) = (st[rkvs1][rk, val][rk0, val0]), rk ∈ rkvs1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]