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Proof step 20 from proof refin-del (previous step)

r ∈ st ⊦ ∃ sk. r -- sk ∈ st -- r1, r = r1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st
Proof step 19:
r ∈ st ⊦ r = r1, ∃ sk. r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]