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Proof step 2 from proof del-size-01 (previous step)

a ∈ s ⊦ # s = n + 1 ↔ #(s -- a) = n

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-size: ⊦ a ∈ s → #(s -- a) = # s - 1;% ⊦ m + n - m = n; (s)% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)