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Proof step 2 from proof sub-succ (previous step)

m0 ≤ n, n ≤ m ⊦ m - (n - m0) = (m - n) + m0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: le-def: ⊦ m ≤ n ↔ ¬ n < m;%le-ls-simp: ⊦ m ≤ n → (n < m ↔ false);%eq-02: n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1;%ls-03: n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0;%fle: ⊦ m < n ∧ n ≤ n0 → m < n0;%sub-add: ⊦ n ≤ n0 → (m + (n0 - n)) + n = m + n0;% ⊦ m + n = m + n0 ↔ n = n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic