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Proof step 2 from proof sw (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 3:
⊦ a0 = a, a = b, (a0 ∈ st[b, d] -- a ↔ a0 ∈ (st -- a)[b, d]) ∧ (st[b, d] -- a)[a0] = (st -- a)[b, d][a0]

Rule argument

fma: a0 = a

Additional information

Used simplifier rules: Del-notin: ⊦ ¬ a ∈ st -- a;%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%lookupisempty: ⊦ (st -- a)[a] = (∅)[a];%At-other: ⊦ a ≠ b → st[b, d][a] = st[a]; Used formulas: a0 = a