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Proof step 12 from proof in-list-subset (previous step)

sks0 ≠ [], sks0 ≠ [], st ⊆ st0, r -- sk ∈ st ∧ st[r -- sk] .refval -- sks0 ∈ st, ∀ r0, st1, st2. r0 -- sks0 ∈ st1 ∧ st1 ⊆ st2 → r0 -- sks0 ∈ st2 ⊦ sks0 = [], r -- sk ∈ st0 ∧ st0[r -- sk] .refval -- sks0 ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ r -- sk ∈ st → r ∈ st
Proof step 11:
st ⊆ st0, r -- sk ∈ st, st[r -- sk] .refval -- sks0 ∈ st, r -- sk ∈ st0, r ∈ st, r ∈ st0, ∀ r0, st1, st2. r0 -- sks0 ∈ st1 ∧ st1 ⊆ st2 → r0 -- sks0 ∈ st2 ⊦ sks0 = [], st0[r -- sk] .refval -- sks0 ∈ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ st ∧ ¬ rk ∈ st0 → ¬ st ⊆ st0])]]