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Proof step 1 from proof ls-pred (conclusion)

⊦ n + 1 < m → m - n - 1 < m

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: lels-01: ⊦ ¬ m ≤ n ↔ n < m;%ls-02: n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n;%succ-01: ⊦ m + 1 ≤ n ↔ m < n;%lels: ⊦ ¬ m < n ↔ n ≤ m;%fle: ⊦ m < n ∧ n ≤ n0 → m < n0;%ls-01: n0 ≤ n ⊦ n - n0 < m ↔ n < m + n0;%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%l-05: ⊦ m + m0 < n → ¬ n + n0 < m; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%trans-1: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic1%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1