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

rk1 ≠ rk, ∀ rk, rk0, st, val. rk = rk0 ∨ (st[rk, val][rkvs0][rk0]) = (st[rkvs0][rk0]) ⊦ rk = rk0, (st[rk1, val0][rk, val][rkvs0][rk0]) = (st[rk1, val0][rkvs0][rk0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ rk, rk0, st, val. rk = rk0 ∨ (st[rk, val][rkvs0][rk0]) = (st[rkvs0][rk0]) ⊦ rk1 = rk, (st[rk1, val0][rk, val][rkvs0][rk0]) = (st[rk1, val0][rkvs0][rk0]), rk = rk0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]