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Proof step 8 from proof del-notin-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ r -- sk ∈ st → r ∈ st
Proof step 7:
⊦ (st -- r0)[r0 -- sk] = ∅[r0 -- sk], r0 ∈ st, r0 -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]])]]