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Proof step 3 from proof at-put-other-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sk ≠ sk0 → (st[r -- sk, val][r0 -- sk0]) = st[r0 -- sk0]], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sks + [] = sks])]]