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Proof step 11 from proof refin-del (previous step)

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

User applied rule prop simplification

And obtained the following new goals

Proof step 10:
¬ r1 ∈ st -- r1, ∀ r0, sk. r1 ≠ r0 → (r0 -- sk ∈ st ↔ r0 -- sk ∈ st -- r1) ∧ st[r0 -- sk] = (st -- r1)[r0 -- sk], r ∈ st -- r1 ⊦ r ∈ st, r = r1

Rule argument

[fmapos-arg: pos: left 1]

Additional information

[[CONCRETESIMPRULES : ()]]