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Proof step 3 from proof del-ne (previous step)

a0 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ # s = 0, # s = 1

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 4:
1 < # s ⊦ ∃ a, b. a ≠ b ∧ a ∈ s ∧ b ∈ s
Proof step 5:
a0 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s, ∃ a, b. a ≠ b ∧ a ∈ s ∧ b ∈ s ⊦ # s = 1, # s = 0

Rule argument

spec: set-basic name: size-two-03 seq: 1 < # s ⊦ ∃ a, b. a ≠ b ∧ a ∈ s ∧ b ∈ s subst: [s] → [s]

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) Used terms: s