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Proof step 11 from proof diff-sub-size (previous step)

a ∈ s0, a0 ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0, (∀ a. a ∈ s → a ∈ s0) → #(s0 \ s) = # s0 - # s ⊦ #(s0 \ s) - 1 = # s0 - # s + 1, a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information