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Proof step 6 from proof dups-eqsize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# x < # y, ∀ a. a ∈ x ↔ a ∈ y ⊦ # x = # y, y = [], dups(x), dups(y)

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat