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Proof step 17 from proof butmin-put-01 (previous step)

a = st[a, d0].min ⊦ a0 = a, a0 = st[a, d].min, a ∈ st, (a0 ∈ st[a, d] -- st[a, d].min ↔ a0 ∈ st) ∧ st[a, d][a0] = st[a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
a = st[a, d0].min ⊦ a0 = a, a0 = st[a, d].min, a0 ∈ st[a, d] -- st[a, d].min ↔ a0 ∈ st, a ∈ st

Rule argument

Additional information

Used simplifier rules: At-other: ⊦ a ≠ b → st[b, d][a] = st[a]; from specification store