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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, m = 0

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, m = 0

Rule argument

pos: left 2

Additional information

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