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

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

Rule argument

Additional information

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