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Proof step 11 from proof lookup-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
st ⊆ st0, r -- sk ∈ st, st[r -- sk] .refval -- sks0 ∈ st, r ∈ st, ∀ r0, st1, st2. r0 -- sks0 ∈ st1 ∧ st1 ⊆ st2 → st2[r0 -- sks0] = st1[r0 -- sks0] ⊦ sks0 = [], st0[st0[r -- sk] .refval -- sks0] = st[st[r -- sk] .refval -- sks0]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]