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Proof step 3 from proof look-del (previous step)

¬ r ∈ st -- r ∧ (∀ r0, sk. r ≠ r0 → (r0 -- sk ∈ st ↔ r0 -- sk ∈ st -- r) ∧ st[r0 -- sk] = (st -- r)[r0 -- sk]) ⊦ r = r0, (st -- r)[r0 -- sk0] = st[r0 -- sk0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ r0, sk. r ≠ r0 → (r0 -- sk ∈ st ↔ r0 -- sk ∈ st -- r) ∧ st[r0 -- sk] = (st -- r)[r0 -- sk] ⊦ (st -- r)[r0 -- sk0] = st[r0 -- sk0], r = r0, r ∈ st -- r

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]