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Proof step 15 from proof in-list-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
st ⊆ st0, r -- (sk ' + sks0) ∈ st, ∀ r0, st1, st2. r0 -- sks0 ∈ st1 ∧ st1 ⊆ st2 → r0 -- sks0 ∈ st2 ⊦ sks0 = [], r -- (sk ' + sks0) ∈ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]