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Proof step 2 from proof in-del (conclusion)
⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 1
:
⊦ r -- sk ∈ st -- r0 ↔ ¬ ¬ (r ≠ r0 ∧ r -- sk ∈ st)
Rule argument
[emptyarg]
Additional information
[[CONCRETESIMPRULES : ()]]