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Proof step 4 from proof del-size-02 (previous step)

a0 ∈ s, a ∈ s, b ∈ s ⊦ #(s -- b -- a) = #(s -- b) - 1, a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
b ∈ s, a ∈ s, a0 ∈ s ⊦ #(s -- b -- a) = # s - 1 - 1, a = b

Rule argument

Additional information

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