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Proof step 6 from proof rmdup-01 (previous step)

a0 ∈ x, ∀ a0. a0 ∈ x → a0 ∈ rmdup(a ' + y0), ∀ x. (∀ a. a ∈ x → a ∈ rmdup(y0)) ↔ (∀ a. a ∈ x → a ∈ y0) ⊦ a0 ∈ a ' + y0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a0

Additional information

Used simplifier rules: in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%rm-in: ⊦ a ∈ rmdup(x + y) ↔ a ∈ x ∨ a ∈ y; from specification list-dup%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list Used terms: a0