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Proof step 8 from proof dups-eqsize-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# x0 = # y0 + # z, ∀ y0, a1. # x0 = # y0 ∧ a1 ∈ y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → a1 ∈ x0 ∨ dups(y0), a ∈ y0 ∨ a ∈ z, ∀ a1. a1 ∈ x0 → a1 ∈ y0 ∨ a1 ∈ z ⊦ a = a0, a0 ∈ x0, dups(x0), a ∈ x0, a0 ∈ y0, a0 ∈ z, dups(y0 + z)

Rule argument

Additional information

Used simplifier rules: f-01: ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y);%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ 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%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%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