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Proof step 18 from proof rm-allsize (previous step)

¬(¬ a ∈ y0 ∧ ¬ a ∈ z), ∀ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
a ∈ z, ∀ y. (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦ # rmdup(x0) + 1 = # rmdup(y0 + z), a ∈ y0, a ∈ x0

Rule argument

Additional information

Used simplifier rules: rm-size-02: ⊦ a ∈ y → # rmdup(x + a ' + y) = # rmdup(x + y);%a: ⊦ (x + y) + z = x + y + z; from specification list