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Proof step 6 from proof subset-put-02 (previous step)

a0 ∈ st, a ∈ st0, a ∈ st[a, st0[a]], ∀ a0. a0 ∈ st[a, st0[a]] → a0 ∈ st0 ∧ st[a, st0[a]][a0] = st0[a0] ⊦ a0 = a, a0 ∈ st0 ∧ st[a0] = st0[a0]

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 7:
a0 ∈ st, a ∈ st0, a ∈ st[a, st0[a]], ∀ a0. a0 ∈ st[a, st0[a]] → a0 ∈ st0 ∧ st[a, st0[a]][a0] = st0[a0] ⊦ a0 = a, a0 = a, a0 ∈ st0 ∧ st[a0] = st0[a0]

Rule argument

fma: a0 = a

Additional information

Used formulas: a0 = a