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Proof step 1 from proof size-del-leq (conclusion)

⊦ # s ≤ #(s -- a) ↔ ¬ a ∈ s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-size: ⊦ a ∈ s → #(s -- a) = # s - 1;%del-in: ⊦ ¬ a ∈ s → s -- a = s;%ler: ⊦ m ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%ls-02: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n; from specification nat%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%size-zero-02: ⊦ # s = 0 → s = ∅; (forward) from specification set-basic%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat