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Proof step 13 from proof rm-allsize (previous step)

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

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 14:
⊦ a ∈ x → (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y)
Proof step 15:
a ∈ y → (∃ y0, z. y = y0 + a ' + z ∧ ¬ a ∈ y0), a ∈ y, ∀ y. (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y), ∀ a0. a0 = a ∨ a0 ∈ x0 ↔ a0 ∈ y ⊦ # rmdup(x0) + 1 = # rmdup(y), a ∈ x0

Rule argument

spec: list inst: name: firstoc seq: ⊦ a ∈ x → (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y) subst: [a, x] → [a, y] flags:

Additional information

Used terms: a, y