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Proof step 8 from proof subset-put (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 9:
a0 ∈ st[a, st0[a]], a ∈ st0, ∀ a0. a0 ∈ st -- a → a0 ∈ st0 ∧ st[a0] = st0[a0] ⊦ a0 = a, a0 ∈ st0 ∧ st[a, st0[a]][a0] = st0[a0]

Rule argument

fma: a0 = a

Additional information

Used simplifier rules: At-same: ⊦ st[a, d][a] = d; Used formulas: a0 = a