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Proof step 4 from proof dups-lesize (previous step)

# y < #(a + x0), ∀ a0. a0 ∈ a + x0 → a0 ∈ y, ∀ y. # y < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y) → x0 = [] ∨ dups(x0) ∨ dups(y) ⊦ a + x0 = [], dups(y), dups(a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# y ≤ # x0, ∀ y0. # y0 < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → x0 = [] ∨ dups(y0), ∀ a0. a0 ∈ a ' + x0 → a0 ∈ y ⊦ dups(x0), dups(y), a ∈ x0

Rule argument

Additional information

Used simplifier rules: dups-cons: ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x;% ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)