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Proof step 7 from proof subset-put-01 (previous step)

a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0], ∀ a0. a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0], a ∈ st → a ∈ st0[a, st[a]] ∧ st[a] = d ⊦ (a0 ∈ st → a0 ∈ st0 ∧ st[a0] = st0[a0]) ∧ ¬ a ∈ st, a ∈ st ∧ st[a] = d ∧ (a1 ∈ st -- a → a1 ∈ st0 -- a ∧ st[a1] = st0[a1])

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 8:
a0 = a, a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0], ∀ a0. a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0], a ∈ st → a ∈ st0[a, st[a]] ∧ st[a] = d ⊦ (a0 ∈ st → a0 ∈ st0 ∧ st[a0] = st0[a0]) ∧ ¬ a ∈ st, a ∈ st ∧ st[a] = d ∧ (a1 ∈ st -- a → a1 ∈ st0 -- a ∧ st[a1] = st0[a1])
Proof step 10:
a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0], ∀ a0. a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0], a ∈ st → a ∈ st0[a, st[a]] ∧ st[a] = d ⊦ a0 = a, (a0 ∈ st → a0 ∈ st0 ∧ st[a0] = st0[a0]) ∧ ¬ a ∈ st, a ∈ st ∧ st[a] = d ∧ (a1 ∈ st -- a → a1 ∈ st0 -- a ∧ st[a1] = st0[a1])

Rule argument

fma: a0 = a

Additional information

Used formulas: a0 = a