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Proof step 23 from proof at-put-other-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ sk ∈ sk0 ' + sks ↔ sk ≠ sk0 ∧ ¬ sk ∈ sks], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sk ' + sks ≠ []], [CSIMPSEQ : ⊦ sk + sks = sk ' + sks])]]