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Proof step 3 from proof dups-eqsize-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# x0 + 1 = # y, ∀ y0, a1. # x0 = # y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → ¬ dups(y0) ∧ (a1 ∈ y0 → a1 ∈ x0), ∀ a1. a1 = a0 ∨ a1 ∈ x0 → a1 ∈ y ⊦ a0 ∈ x0, dups(x0), ¬ dups(y) ∧ (a ∈ y → a = a0 ∨ a ∈ x0)

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%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list