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Proof step 9 from proof min-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
⊦ # st.keys = 0, st.keys = ∅, a = (st.keys ++ a).min, (st.keys ++ a).min = st.min, a < st.keys.min

Rule argument

Additional information

Used simplifier rules: size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic