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Proof step 15 from proof lookup-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ r -- sk ∈ st → r ∈ st
Proof step 14:
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] ⊦ st0[r -- (sk ' + sks0)] = st[r -- (sk ' + sks0)], sks0 = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]