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Proof step 3 from proof union-size-le (previous step)

# s0 + #(s ++ a) < #(s0 ∪ (s ++ a)), ∀ s0. ¬ # s0 + # s < #(s0 ∪ s) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# s0 + #(s ++ a) < #((s ∪ s0) ++ a), ∀ s0. ¬ # s0 + # s < #(s0 ∪ s) ⊦

Rule argument

Additional information

Used simplifier rules: unadd: ⊦ (s ++ a) ∪ s0 = (s ∪ s0) ++ a;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic