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Proof step 4 from proof dups-eqsize-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
1 + # x0 = # y, a ∈ y, ∀ y0, a1. # x0 = # y0 ∧ a1 ∈ y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → dups(y0) ∨ a1 ∈ x0, ∀ a1. a1 = a0 ∨ a1 ∈ x0 → a1 ∈ y ⊦ a = a0, dups(y), dups(x0), a ∈ x0, a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: dups-cons: ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x;% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a ∈ b ' + x ↔ ¬(a ≠ b ∧ ¬ a ∈ x); (ss)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)