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Proof step 5 from proof dups-lesize (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 6:
# y ≤ # x0, a ∈ a ' + x0 → a ∈ y, ∀ a0. a0 ∈ a ' + x0 → a0 ∈ y, ∀ y0. # y0 < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → x0 = [] ∨ dups(y0) ⊦ dups(y), dups(x0), a ∈ x0

Rule argument

pos: left 3 terms: a abort: F discard: F computed: T

Additional information

Used terms: a