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Proof step 6 from proof at-put-other-02 (previous step)

sks0 ≠ [], ∀ st, r, sk, val, r0. (st[r -- sk, val][r0 -- sks0]) = st[r0 -- sks0] ∨ sk ∈ sks0 ⊦ sk = sk0, sks0 = [], (st[r -- sk, val][st[r0 -- sk0] .refval -- sks0]) = st[st[r0 -- sk0] .refval -- sks0], sk ∈ sks0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ st, r, sk, val, r0. (st[r -- sk, val][r0 -- sks0]) = st[r0 -- sks0] ∨ sk ∈ sks0 ⊦ sks0 = [], (st[r -- sk, val][st[r0 -- sk0] .refval -- sks0]) = st[st[r0 -- sk0] .refval -- sks0], sk = sk0, sk ∈ sks0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]