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Proof step 5 from proof q-01 (previous step)

m0 < n, n < m0 ∨ (∃ m1. m0 + m1 = n) ⊦ n = 0, ∃ m. (m0 + 1) + m = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ m0 + m = 0, m = 0, ∃ m1. 1 + m1 = m

Rule argument

Additional information

Used simplifier rules: l-01: ⊦ m < m + n ↔ n ≠ 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%eq-01: ⊦ m + n = m + n0 ↔ n = n0;%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1