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Proof step 16 from proof rm-allsize (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ ¬ a ∈ x ∧ ¬ a ∈ y → # rmdup(x + a ' + y) = # rmdup(x + y) + 1
Proof step 18:
¬(¬ a ∈ y0 ∧ ¬ 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 + a ' + z), a ∈ y0, a ∈ x0
Proof step 41:
¬ a ∈ y0 ∧ ¬ 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) + 1, a ∈ y0, a ∈ x0

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] → [z, a, y0] flags: rewrite, current

Additional information

Used terms: z, a, y0