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Proof step 5 from proof dups-01 (previous step)

b ∈ (a ' + y0) -1l b, ∀ b0. ¬ b0 ∈ y0 -1l b0 ∨ ¬ b0 ∈ y0, b = a ∨ b ∈ y0 ⊦ b = a, dups(y0), a ∈ y0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
b ∈ y0 -1l b, b ∈ y0, ∀ b0. ¬ b0 ∈ y0 -1l b0 ∨ ¬ b0 ∈ y0 ⊦ dups(y0), a ∈ y0

Rule argument

Additional information

Used simplifier rules: del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a);%in-01-01: ⊦ a ∈ x -1l b → a ∈ x;%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list