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Proof step 5 from proof del-three (previous step)

a ∈ s, #(s -- a) - 1 = 0, b ∈ s, 2 < # s, a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# s - 1 - 1 = 0, a ∈ s, b ∈ s, 2 < # s, a0 ∈ s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦ b = a

Rule argument

Additional information

Used simplifier rules: del-size: ⊦ a ∈ s → #(s -- a) = # s - 1;