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Proof step 10 from proof in-del-01 (previous step)

¬ r0 ∈ st -- r0 ∧ (∀ r1, sk. r0 ≠ r1 → (r1 -- sk ∈ st ↔ r1 -- sk ∈ st -- r0) ∧ st[r1 -- sk] = (st -- r0)[r1 -- sk]) ⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ r ≠ r0 → (st -- r)[r0 -- sk0] = st[r0 -- sk0]
Proof step 9:
∀ r1, sk0. r0 ≠ r1 → (r1 -- sk0 ∈ st ↔ r1 -- sk0 ∈ st -- r0) ⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st, r0 ∈ st -- r0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]