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Proof step 2 from proof sub-sizeneqex (previous step)

s ⊆ s0 ⊦ # s = # s0, ∃ a. a ∈ s0 ∧ ¬ a ∈ s

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
# s < # s0 ⊦ ∃ a. a ∈ s0 ∧ ¬ a ∈ s

Rule argument

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

Additional information

Used simplifier rules: sub-size: ⊦ s ⊆ s0 → ¬ # s0 < # s;% ⊦ m ≠ n → (m < n ↔ ¬ n < m); (ss) Used terms: s, s0