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Proof step 5 from proof ins3 (previous step)

⊦ (a0 ∈ st[a, d1][b, d0][a, d] ↔ a0 ∈ st[b, d0][a, d]) ∧ st[a, d1][b, d0][a, d][a0] = st[b, d0][a, d][a0]

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 6:
⊦ a0 = a, (a0 ∈ st[a, d1][b, d0][a, d] ↔ a0 ∈ st[b, d0][a, d]) ∧ st[a, d1][b, d0][a, d][a0] = st[b, d0][a, d][a0]

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; Used formulas: a0 = a