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Proof step 4 from proof eq-shift-del (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a0 = a, ∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ (a0 ∈ st[a, d] → st[a, d][a0] = st0[a0]) ⊦ a ∈ st, (a0 ∈ st ↔ a0 ∈ st0 -- a) ∧ (a0 ∈ st → st[a0] = (st0 -- a)[a0]) ∧ a ∈ st0 ∧ st0[a] = d
Proof step 7:
∀ a0. (a0 ∈ st[a, d] ↔ a0 ∈ st0) ∧ (a0 ∈ st[a, d] → st[a, d][a0] = st0[a0]) ⊦ a0 = a, a ∈ st, (a0 ∈ st ↔ a0 ∈ st0 -- a) ∧ (a0 ∈ st → st[a0] = (st0 -- a)[a0]) ∧ a ∈ st0 ∧ st0[a] = d

Rule argument

fma: a0 = a

Additional information

Used formulas: a0 = a