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Proof step 6 from proof union-size-02 (previous step)

a ∈ s0, a0 ∈ s0 ⊦ #(s ∪ s0) = (#(s ∪ s0) - 1) + 1, a ∈ s, 1 ≤ #(s ∪ s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
#(s ∪ s0) = 0, a ∈ s0, a0 ∈ s0, ∀ a1. ¬ a1 ∈ s, ∀ a1. ¬ a1 ∈ s0 ⊦ 0 = (0 - 1) + 1, a ∈ s

Rule argument

Additional information

Used simplifier rules: une-01: ⊦ s1 ∪ s2 = ∅ ↔ s1 = ∅ ∧ s2 = ∅;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)