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Proof step 41 from proof del-exists (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
∀ a0. a0 ∈ st[a, st0[a]] → (∃ st1. ¬ a0 ∈ st1 ∧ st[a, st0[a]] = st1[a0, st[a, st0[a]][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]]

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;%At-same: ⊦ st[a, d][a] = d;