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Proof step 5 from proof delete-put-rew (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: a0 = a

Additional information

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