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Proof step 3 from proof del-size-02 (previous step)

b ∈ s, a ∈ s ⊦ s = ∅, a = b, #(s -- b -- a) = #(s -- b) - 1, # s = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
b ∈ s, a ∈ s ⊦ # s = 0, s = ∅, a = b, #(s -- b -- a) = # s - 1 - 1

Rule argument

Additional information

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