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Proof step 5 from proof del-exists (previous step)

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]]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ ∃ st. ¬ a0 ∈ st ∧ st0[a0, d] = st[a0, d]

Rule argument

Additional information

Used simplifier rules: At-same: ⊦ st[a, d][a] = d;