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Proof step 29 from proof del-exists (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 2 flags: discard

Additional information

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