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

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

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], a ∈ st → a ∈ st0[a, st[a]] ∧ st[a] = d, ∀ a0. a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0] ⊦ a ∈ st ∧ st[a] = d ∧ (a0 ∈ st -- a → a0 ∈ st0 -- a ∧ (st -- a)[a0] = (st0 -- a)[a0]), (a1 ∈ st → a1 ∈ st0 ∧ st[a1] = st0[a1]) ∧ ¬ a ∈ st
Proof step 11:
a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0], a ∈ st → a ∈ st0[a, st[a]] ∧ st[a] = d, ∀ a0. a0 ∈ st → a0 ∈ st0[a, d] ∧ st[a0] = st0[a, d][a0] ⊦ a0 = a, a ∈ st ∧ st[a] = d ∧ (a0 ∈ st -- a → a0 ∈ st0 -- a ∧ (st -- a)[a0] = (st0 -- a)[a0]), (a1 ∈ st → a1 ∈ st0 ∧ st[a1] = st0[a1]) ∧ ¬ a ∈ st

Rule argument

fma: a0 = a

Additional information

Used formulas: a0 = a