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Proof step 13 from proof union-size-le (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
# s0 + # s < #(s ∪ s0), a ∈ s, a0 ∈ s, ∀ s0. ¬ # s0 + # s < #(s0 ∪ s) ⊦

Rule argument

Additional information

Used simplifier rules: uninsert: ⊦ a ∈ s → (s ∪ s0) ++ a = s ∪ s0;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)% ⊦ m < n → n ≠ 0; (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic