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Proof step 25 from proof rm-allsize (previous step)

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 + a ' + z0), a ∈ y0, a ∈ x0, a ∈ y, a ∈ z0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 26:
⊦ ¬ a ∈ x ∧ ¬ a ∈ y → # rmdup(x + a ' + y) = # rmdup(x + y) + 1
Proof step 27:
¬ 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

Rule argument

spec: list-dup inst: name: rm-size seq: ⊦ ¬ a ∈ x ∧ ¬ a ∈ y → # rmdup(x + a ' + y) = # rmdup(x + y) + 1 subst: [y, a, x] → [z0, a, y0 + y] flags: rewrite, current

Additional information

Used simplifier rules: in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: z0, a, y0 + y