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Proof step 4 from proof dups-01 (previous step)

b ∈ a + y0, b ∈ (a + y0) -1l b, ∀ b. b ∈ y0 -1l b ∧ b ∈ y0 → dups(y0) ⊦ dups(a + y0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
b ∈ (a ' + y0) -1l b, ∀ b0. ¬ b0 ∈ y0 -1l b0 ∨ ¬ b0 ∈ y0, b = a ∨ b ∈ y0 ⊦ dups(y0), a ∈ y0

Rule argument

Additional information

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