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Proof step 2 from proof ls-pred (previous step)

n + 1 < m ⊦ m - n - 1 < m

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ 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;%succ-02: ⊦ m < n + 1 ↔ m ≤ n;% ⊦ ¬ m + n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic