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Proof step 19 from proof rm-allsize (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 20:
⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z))
Proof step 21:
a ∈ rmdup(z) ↔ (∃ y, z0. rmdup(z) = y + (a + z0)), a ∈ z, ∀ 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), a ∈ x0, a ∈ y0

Rule argument

spec: list name: in-def seq: ⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z)) subst: [a, x] → [a, rmdup(z)]

Additional information

Used terms: a, rmdup(z)