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Proof step 1 from proof del-ne (conclusion)

⊦ 1 < # s → s -- a ≠ ∅

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a;%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%one-size: ⊦ s = ∅ ++ a → # s = 1; (forward) from specification set-basic%size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1