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Proof step 3 from proof q-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m0 < n, ∀ n. n < m0 ∨ (∃ m1. m0 + m1 = n) ⊦ n = 0, ∃ m. (m0 + 1) + m = n

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%s: ⊦ m < n + 1 ↔ ¬ n < m;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1