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Proof step 4 from proof del-put (previous step)

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

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: a0 = a

Additional information

Used simplifier rules: At-same: ⊦ st[a, d][a] = d;%In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;%At-other: ⊦ a ≠ b → st[b, d][a] = st[a];%Del-at: ⊦ a ≠ b → (st -- b)[a] = st[a]; Used formulas: a0 = a