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Proof step 17 from proof in-list-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
st ⊆ st0, r -- (sk ' + sks0) ∈ st, ∀ r, st, st0. r -- sks0 ∈ st ∧ st ⊆ st0 → sks0 = [] ∨ r -- sks0 ∈ st0 ⊦ r -- (sk ' + sks0) ∈ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sk ' + sks ≠ []], [CSIMPSEQ : ⊦ sk + sks = sk ' + sks])]]