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Proof step 2 from proof size-two-04 (previous step)

a ∈ s ⊦ # s = 0, # s = 1, ∃ b. b ∈ s ∧ a ≠ b

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s
Proof step 4:
∃ a0. a0 ∈ s ∧ ¬ a0 ∈ ∅ ++ a, a ∈ s ⊦ # s = 0, # s = 1, ∃ b. b ∈ s ∧ a ≠ b

Rule argument

spec: set-basic inst: name: size-exgr seq: # s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s subst: [s, s0] → [∅ ++ a, s] flags: current, []

Additional information

Used simplifier rules: size-one-01: ⊦ #(∅ ++ a) = 1;%one-02: ⊦ m ≠ 0 → (¬ 1 < m ↔ m = 1); from specification nat-basic1 Used terms: ∅ ++ a, s