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Proof step 7 from proof min-put (previous step)

¬(a < st.keys.min ∧ st.keys ≠ ∅) ⊦ (st.keys ++ a).min = st.min, a = (st.keys ++ a).min

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ (st.keys ++ a).min = st.min, a = (st.keys ++ a).min, a < st.keys.min ∧ st.keys ≠ ∅

Rule argument

Additional information