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Proof step 16 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 ⊦ r -- (sk ' + sks0) ∈ st0

User applied rule cut formula

And obtained the following new goals

Proof step 1:
⊦ r -- sk ' ∈ st ↔ r -- sk ∈ st
Proof step 15:
st ⊆ st0, r -- (sk ' + sks0) ∈ st, ∀ r, st, st0. r -- sks0 ∈ st ∧ st ⊆ st0 → sks0 = [] ∨ r -- sks0 ∈ st0 ⊦ sks0 = [], r -- (sk ' + sks0) ∈ st0

Rule argument

[fma-arg: sks0 = []]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sks + [] = sks], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ rk ∈ st ∧ ¬ rk ∈ st0 → ¬ st ⊆ st0])], [EXTRAFMAS : (sks0 = [])]]