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Proof step 7 from proof delete-put-rew (previous step)

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

Heuristic weak cut applied rule cut formula

And closed the goal!

Rule argument

fma: a0 = b

Additional information

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