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Proof step 3 from proof diff-sub-size (previous step)

∀ a. a ∈ ∅ → a ∈ s0 ⊦ #(s0 \ ∅) = # s0 - # ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ # s0 = # s0 - 0

Rule argument

Additional information

Used simplifier rules: diff-e: ⊦ s \ ∅ = s;% ⊦ ¬ a ∈ ∅; (s)% ⊦ # ∅ = 0; (s)