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Proof step 2 from proof sw-01 (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

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

Rule argument

fma: a0 = b

Additional information

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