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Proof step 8 from proof dups-lesize (previous step)

# y ≤ # x0, ∃ y0, z. y = y0 + (a + z), ∀ a0. a0 ∈ x0 → a0 ∈ y, ∀ y0. # y0 < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → x0 = [] ∨ dups(y0) ⊦ a ∈ x0, dups(x0), dups(y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# y0 + # z < # x0, ∀ a0. a0 ∈ x0 → a0 ∈ y0 ∨ a0 ∈ z, ∀ y. # y < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y) → dups(y) ⊦ x0 = [], dups(y0 + z), a ∈ z, a ∈ y0, dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: f-01: ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y);%l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list