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Proof step 22 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 + z), a ∈ y0, a ∈ x0, a ∈ y, a ∈ z0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ rmdup(x + rmdup(y)) = rmdup(x + y)
Proof step 24:
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 + rmdup(z)), a ∈ y0, a ∈ x0, a ∈ y, a ∈ z0

Rule argument

spec: list-dup inst: name: rm-rm-02 seq: ⊦ rmdup(x + rmdup(y)) = rmdup(x + y) subst: [y, x] → [z, y0] flags: rewrite, current

Additional information

Used terms: z, y0