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Proof step 10 from proof eq-shift-put (previous step)

a ∈ st, ∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ st[a, d][a0] = st0[a0] ⊦ (a0 ∈ st ↔ a0 ∈ st0[a, st[a]]) ∧ st[a0] = st0[a, st[a]][a0] ∧ a ∈ st0 ∧ st0[a] = d

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 11:
a0 = a, a ∈ st, ∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ st[a, d][a0] = st0[a0] ⊦ (a0 ∈ st ↔ a0 ∈ st0[a, st[a]]) ∧ st[a0] = st0[a, st[a]][a0] ∧ a ∈ st0 ∧ st0[a] = d
Proof step 13:
a ∈ st, ∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ st[a, d][a0] = st0[a0] ⊦ a0 = a, (a0 ∈ st ↔ a0 ∈ st0[a, st[a]]) ∧ st[a0] = st0[a, st[a]][a0] ∧ a ∈ st0 ∧ st0[a] = d

Rule argument

fma: a0 = a

Additional information

Used formulas: a0 = a