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Proof step 3 from proof size-two-04 (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s
Proof step 5:
a ∈ s, ∃ a0. a0 ∈ s ∧ ¬ a0 ∈ ∅ ++ a ⊦ # s = 0, # s = 1, ∃ b. b ∈ s ∧ a ≠ b

Rule argument

spec: <toplevel> name: size-exgr seq: # s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s subst: [s, s0] → [∅ ++ a, s]

Additional information

Used simplifier rules: size-one-01: ⊦ #(∅ ++ a) = 1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s) Used terms: ∅ ++ a, s