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Proof step 18 from proof in-list-subset (previous step)

r -- sks ∈ st, st ⊆ st0 ⊦ sks = [], r -- sks ∈ st0

User applied rule structural induction

And obtained the following new goals

Proof step 17:
∀ r, st, st0. r -- sks0 ∈ st ∧ st ⊆ st0 → sks0 = [] ∨ r -- sks0 ∈ st0, r -- (sk + sks0) ∈ st, st ⊆ st0 ⊦ sk + sks0 = [], r -- (sk + sks0) ∈ st0

Rule argument

[VARWITHVARSEQSARG : sks, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : storekeys generated by [], +;])]]