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Proof step 18 from proof refin-del (
previous step
)
∃ sk. r -- sk ∈ st ⊦ r = r1, ∃ sk. r -- sk ∈ st
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 16
:
⊦ r -- sk ∈ st → r ∈ st
Proof step 17
:
r -- sk ∈ st, r ∈ st ⊦ r = r1, ∃ sk. r -- sk ∈ st
Rule argument
[emptyarg]
Additional information
[[CONCRETESIMPRULES : ()]]