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

⊦ r ∈ st, (r0 -- sk ∈ st -- r ↔ r0 -- sk ∈ st) ∧ (st -- r)[r0 -- sk] = st[r0 -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]