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Proof step 8 from proof sub-size-eq (previous step)

a ∈ s0, # s0 ≤ # s, s ⊆ s0 ⊦ a ∈ s

User applied rule insert lemma

And obtained the following new goals

Proof step 9:
⊦ s ⊆ s0 ∧ a ∈ s0 ∧ ¬ a ∈ s → # s < # s0

Rule argument

spec: <toplevel> name: sizesubl seq: ⊦ s ⊆ s0 ∧ a ∈ s0 ∧ ¬ a ∈ s → # s < # s0 subst: [s, s0, a] → [s, s0, a]

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s) Used terms: s, s0, a