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

# s0 = 0 → s0 = ∅, #(s0 ++ a) = 0 ⊦ s0 ++ a = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
#(s0 ++ a) = 0 ⊦ s0 = ∅, # s0 = 0

Rule argument

Additional information

Used simplifier rules: ne: ⊦ ∅ ≠ s ++ a;%size-one-01: ⊦ #(∅ ++ a) = 1;%size-zero: ⊦ # s ≠ 0 → s ≠ ∅;