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Proof step 2 from proof dups-eqsize (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
¬ dups(x), ¬ dups(y), ∀ a. a ∈ x → a ∈ y ⊦ # x ≤ # y
Proof step 4:
∀ a. a ∈ x ↔ a ∈ y ⊦ ¬ dups(x) ∧ ¬ dups(y) ∧ (∀ a. a ∈ x → a ∈ y), # x = # y, dups(y), dups(x)
Proof step 6:
# x ≤ # y, ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y, dups(y), dups(x)

Rule argument

spec: list-dup inst: name: dups-lesize seq: ¬ dups(x), ¬ dups(y), ∀ a. a ∈ x → a ∈ y ⊦ # x ≤ # y subst: [x, y] → [x, y] flags: current, []

Additional information

Used terms: x, y