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Proof step 9 from proof dups-eqsize-01 (previous step)

1 + # x0 = # y, a ∈ y, ∀ a1. a1 ∈ x0 → a1 ∈ y, ∀ y0, a1. # x0 = # y0 ∧ a1 ∈ y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → dups(y0) ∨ a1 ∈ x0, ∃ y0, z. y = y0 + (a0 + z) ⊦ a = a0, dups(x0), dups(y), a0 ∈ x0, a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# x0 = # y0 + # z, ∀ y0, a1. # x0 = # y0 ∧ a1 ∈ y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → dups(y0) ∨ a1 ∈ x0, ∀ a1. a1 ∈ x0 → a1 ∈ y0 ∨ a1 ∈ z, a ∈ y0 ∨ a ∈ z ⊦ a = a0, dups(x0), dups(y0 + z), a ∈ x0, a0 ∈ x0, a0 ∈ y0, a0 ∈ z

Rule argument

Additional information

Used simplifier rules: f-01: ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y);% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)