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Proof step 3 from proof del-three (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
#(s -- a) - 1 = 0, 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;%Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic