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Proof step 8 from proof del-exists (previous step)

∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]), a0 ∈ st0 → ¬ a0 ∈ st ∧ st0 = st[a0, st0[a0]] ⊦ ∃ st. ¬ a0 ∈ st ∧ st0[a0, d] = st[a0, d]

User applied rule cut formula

And obtained the following new goals

Proof step 9:
a0 ∈ st0, ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]), a0 ∈ st0 → ¬ a0 ∈ st ∧ st0 = st[a0, st0[a0]] ⊦ ∃ st. ¬ a0 ∈ st ∧ st0[a0, d] = st[a0, d]

Rule argument

fma: a0 ∈ st0

Additional information

Used simplifier rules: inj-put: ⊦ ¬ a ∈ st ∧ ¬ a ∈ st0 → (st[a, d] = st0[a, d] ↔ st = st0); Used formulas: a0 ∈ st0