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Proof step 4 from proof indef (previous step)

a ∈ s ⊦ ∃ s0. (∀ a0. a0 ∈ s ↔ a0 = a ∨ a0 ∈ s0) ∧ ¬ a ∈ s0

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
a ∈ s ⊦ ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0
Proof step 6:
a ∈ s, ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 ⊦ ∃ s0. (∀ a0. a0 ∈ s ↔ a0 = a ∨ a0 ∈ s0) ∧ ¬ a ∈ s0

Rule argument

spec: <toplevel> name: del seq: a ∈ s ⊦ ∃ s0. s = s0 ++ a ∧ ¬ a ∈ s0 subst: [a, s] → [a, s]

Additional information

Used terms: a, s