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Proof step 2 from proof del-three (previous step)

#(s -- a -- b) = 0, 2 < # s ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
#(s -- a -- b) = 0, 2 < # s, ∀ a0. a0 = b ∨ a0 = a ∨ ¬ a0 ∈ s ⊦

Rule argument

Additional information

Used simplifier rules: Delete: ⊦ a ∈ s -- b ↔ a ≠ b ∧ a ∈ s;% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic