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Proof step 4 from proof elim-sub (previous step)

n1 < m, n1 ≤ m - 1 → (n0 = m - 1 - n1 ↔ m - 1 = n0 + n1) ⊦ n0 = m - n1 + 1 ↔ m = n0 + n1 + 1

User applied rule case distinction

And obtained the following new goals

Proof step 5:
n0 = m - 1 - n1 ↔ m - 1 = n0 + n1, n1 < m ⊦ n0 = m - n1 + 1 ↔ m = n0 + n1 + 1

Rule argument

pos: left 2

Additional information

Used simplifier rules: le-def: ⊦ m ≤ n ↔ ¬ n < m;%num-01: ⊦ n + 1 - 1 = n;%succ-02: ⊦ m < n + 1 ↔ m ≤ n;%num-02: n ≠ 0 ⊦ (n - 1) + 1 = n;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m +1 = m + 1; (s)% ⊦ ¬ m + n < m; (s)% ⊦ ¬ n < n; (s)