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Proof step 27 from proof diff-sub-size (previous step)

# s0 = #(s0 \ s) + # s, a0 ∈ s0, a ∈ s0, # s ≤ # s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0 ⊦ #(s0 \ s) - 1 = # s0 - # s + 1, a ∈ s, # s + 1 ≤ # s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
#(s0 \ s) = 0, # s = # s0, a ∈ s0, a0 ∈ s0, ∀ a0. a0 ∈ s ++ a → a0 ∈ s0, ∀ a1. ¬ a1 ∈ s0 \ s ⊦ 0 - 1 = # s0 - # s0 + 1, a ∈ s

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic% ⊦ m ≤ m; (s)% ⊦ m = m + n ↔ n = 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m + 1 ≤ n ↔ m < n; (s)