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Proof step 9 from proof at-put-other-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ st[r -- sk '] = st[r -- sk]
Proof step 8:
⊦ sk = sk0, (st[r -- sk, val][jvmref -- sk0]) = st[jvmref -- sk0]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ sk ∈ []], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sks + [] = sks])]]