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Proof step 15 from proof rm-allsize (previous step)

a ∈ y → (∃ y0, z. y = y0 + a ' + z ∧ ¬ a ∈ y0), a ∈ y, ∀ y. (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y), ∀ a0. a0 = a ∨ a0 ∈ x0 ↔ a0 ∈ y ⊦ # rmdup(x0) + 1 = # rmdup(y), a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
∀ y. (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦ # rmdup(x0) + 1 = # rmdup(y0 + a ' + z), a ∈ y0, a ∈ x0

Rule argument

Additional information

Used simplifier rules: in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%in-13: ⊦ x = y + a ' + z → a ∈ x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list