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Proof step 4 from proof del-exists (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
a = a0, ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ ∃ st1. ¬ a ∈ st1 ∧ st0[a0, d] = st1[a, st0[a0, d][a]]
Proof step 13:
a ∈ st0, ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ ∃ st1. ¬ a ∈ st1 ∧ st0[a0, d] = st1[a, st0[a0, d][a]]

Rule argument

pos: left 2

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