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Proof step 9 from proof in-butmin (previous step)

st .min = r0 -- sk, ∃ sk. r -- sk ∈ st ⊦ r = r0, r ∈ st .butmin

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ r -- sk ∈ st → r ∈ st
Proof step 8:
st .min = r0 -- sk, r -- sk0 ∈ st, r ∈ st ⊦ r = r0, r ∈ st .butmin

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]