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Proof step 2 from proof size (previous step)

y ⊆ x, x ⊆ y ⊦ # x = # y, dups(x), dups(y)

User applied rule insert spec-lemma

And obtained the following new goals

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

Rule argument

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

Additional information

Used terms: x, y