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Proof step 2 from proof sub-size (previous step)

x ⊆ y ⊦ dups(x), # x ≤ # y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# y < # x, x ⊆ y ⊦ x = [], dups(x)

Rule argument

Additional information

Used simplifier rules: ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)