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Proof step 37 from proof diff-sub-size (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;% ⊦ a ∈ s → s ++ a = s; (s)% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)