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Proof step 41 from proof rm-allsize (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
∀ y. (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦ # rmdup(x0) = # rmdup(y0 + z), a ∈ y0, a ∈ z, a ∈ x0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)