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Proof step 17 from proof at-put-other-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
∀ st, r, sk, val. (st[r -- sk, val][jvmref -- sks0]) = st[jvmref -- sks0] ∨ sk ∈ sks0 ⊦ sks0 = [], sk = sk0, (st[r -- sk, val][(st[r -- sk, val][jvmref -- sk0]) .refval -- sks0]) = st[st[jvmref -- sk0] .refval -- sks0], sk ∈ sks0

Rule argument

[emptyarg]

Additional information

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