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Proof step 6 from proof union-size-le (previous step)

¬ a ∈ s, # s0 + #(s ++ a) < #((s ∪ s0) ++ a), ∀ s0. # s0 + # s < #(s0 ∪ s) → #(s0 ∪ s) = 0 ∨ s0 ∪ s = ∅ ⊦ #((s ∪ s0) ++ a) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# s0 + # s + 1 < #((s ∪ s0) ++ a), ∀ s0. # s0 + # s < #(s0 ∪ s) → #(s0 ∪ s) = 0 ∨ s0 ∪ s = ∅ ⊦ #((s ∪ s0) ++ a) = 0, a ∈ s

Rule argument

Additional information

Used simplifier rules: a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%Size-insert: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1; from specification set-basic