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Proof step 43 from proof del-exists (previous step)

∀ a0. a0 ∈ st[a, st0[a]] → (∃ st1. ¬ a0 ∈ st1 ∧ st[a, st0[a]] = st1[a0, st[a, st0[a]][a0]]) ⊦ a1 = a0, a = a0, a ∈ st, (a1 ∈ st[a, st0[a]][a0, d] ↔ a1 ∈ st[a0, d][a, st0[a]]) ∧ st[a, st0[a]][a0, d][a1] = st[a0, d][a, st0[a]][a1], ∃ st1. ¬ a ∈ st1 ∧ st[a, st0[a]][a0, d] = st1[a, st0[a]]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 44:
∀ a0. a0 ∈ st[a, st0[a]] → (∃ st1. ¬ a0 ∈ st1 ∧ st[a, st0[a]] = st1[a0, st[a, st0[a]][a0]]) ⊦ a1 = a0, a = a0, a ∈ st, ∃ st1. ¬ a ∈ st1 ∧ st[a, st0[a]][a0, d] = st1[a, st0[a]], (a1 ∈ st[a, st0[a]] ↔ a1 ∈ st[a0, d][a, st0[a]]) ∧ st[a, st0[a]][a1] = st[a0, d][a, st0[a]][a1]

Rule argument

Additional information

Used simplifier rules: in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st);%At-other: ⊦ a ≠ b → st[b, d][a] = st[a];