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Proof step 2 from proof id-idx-imp-ls-1 (previous step)

n < # lpt1, id-idx(lpt0, lpt1) ⊦ n < # lpt0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: id-idx-ls0: ⊦ id-idx(lpt0, lpt1) → ¬ # lpt0 < # lpt1;% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)