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Proof step 10 from proof dups-lesize (previous step)

# y0 + # z < # x0, ∀ a0. a0 ∈ x0 → a0 ∈ y0 ∨ a0 ∈ z, ∀ y. # y < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y) → dups(y) ⊦ x0 = [], dups(y0 + z), dups(x0), a ∈ z, a ∈ y0, a ∈ x0

User applied rule all left

And obtained the following new goals

Proof step 11:
# y0 + # z < # x0, #(y0 + z) < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y0 + z) → dups(y0 + z), ∀ a0. a0 ∈ x0 → a0 ∈ y0 ∨ a0 ∈ z ⊦ x0 = [], dups(x0), dups(y0 + z), a ∈ x0, a ∈ y0, a ∈ z

Rule argument

pos: left 3 terms: y0 + z abort: F discard: T computed: F

Additional information

Used terms: y0 + z