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Proof step 27 from proof rm-allsize (previous step)

¬ a ∈ y0 + y ∧ ¬ a ∈ z0, rmdup(z) = y + a ' + z0, 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 + y) + z0) + 1, a ∈ y0, a ∈ x0, a ∈ y, a ∈ z0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
rmdup(z) = y + a ' + z0, a ∈ z, ∀ y. (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y), ∀ a0. a0 ≠ a → (a0 ∈ x0 ↔ a0 ∈ y0 ∨ a0 ∈ z) ⊦ # rmdup(x0) = # rmdup((y0 + y) + z0), a ∈ y0, a ∈ x0, a ∈ y, a ∈ z0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2