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Proof step 11 from proof del-three (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
#(s -- a) = 0, 2 < # s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ b ≠ a ∧ b ∈ s

Rule argument

Additional information

Used simplifier rules: del-in: ⊦ ¬ a ∈ s → s -- a = s;%del-rax: ⊦ 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%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic