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Proof step 11 from proof union-size-le (previous step)

# s0 + # s + 1 < #(s ∪ s0), a ∈ s ∪ s0, ∀ s0. ¬ # s0 + # s < #(s0 ∪ s) ⊦ a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# s0 + # s + 1 < #(s ∪ s0), a ∈ s0, a0 ∈ s0, ∀ s0. ¬ # s0 + # s < #(s0 ∪ s) ⊦ a ∈ s

Rule argument

Additional information

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