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Proof step 5 from proof diff-sub-size (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ s0. (∀ a. a ∈ s → a ∈ s0) → #(s0 \ s) = # s0 - # s, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #((s0 \ s) -- a) = # s0 - #(s ++ a)

Rule argument

Additional information

Used simplifier rules: diff-add-del: ⊦ s1 \ (s2 ++ a) = (s1 \ s2) -- a;