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Proof step 8 from proof dups-eqsize (previous step)

# x < # y, ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y, y = [], dups(x), dups(y)

User applied rule insert lemma

And obtained the following new goals

Proof step 9:
¬ dups(x), ¬ dups(y), ∀ a. a ∈ x → a ∈ y ⊦ # x ≤ # y
Proof step 10:
# x < # y, ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y, y = [], dups(y), dups(x), ¬ dups(y) ∧ ¬ dups(x) ∧ (∀ a. a ∈ y → a ∈ x)

Rule argument

spec: <toplevel> name: dups-lesize seq: ¬ dups(x), ¬ dups(y), ∀ a. a ∈ x → a ∈ y ⊦ # x ≤ # y subst: [x, y] → [y, x]

Additional information

Used simplifier rules: ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m < n → ¬ n < m; (s) Used terms: y, x