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Proof step 3 from proof dups-lesize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# y ≤ # x0, ∀ y0. # y0 < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → x0 = [] ∨ dups(y0), ∀ a0. a0 = a ∨ a0 ∈ x0 → a0 ∈ y ⊦ a ∈ x0, dups(x0), dups(y)

Rule argument

Additional information

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