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Proof step 34 from proof del-exists (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
st0 = st[a, st0[a]], a ∈ st0, ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ a = a0, a ∈ st, ∃ st1. ¬ a ∈ st1 ∧ st0[a0, d] = st1[a, st0[a]]

Proof step 34 from proof del-exists (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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