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Proof step 8 from proof rm-allsize (previous step)

a ∈ x0, (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y), ∀ a0. a0 ∈ a ' + x0 ↔ a0 ∈ y, ∀ y. (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y) ⊦ # rmdup(x0) = # rmdup(y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
a ∈ x0, ∀ a0. a0 ∈ x0 ↔ a0 ∈ y, ∀ y. (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y) ⊦ # rmdup(x0) = # rmdup(y), a0 ∈ x0 ↔ a0 ∈ y

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)