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Proof step 10 from proof del-three (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
# s - 1 = 0, 2 < # s, b ∈ s, a0 ∈ s, ∀ a1. a1 = b ∨ ¬ a1 ∈ s ⊦ a ∈ s

Rule argument

Additional information

Used simplifier rules: del-in: ⊦ ¬ a ∈ s → s -- a = s;