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Proof step 4 from proof succ-05 (previous step)

m = n0 + n ∧ n ≤ m, n0 - 1 = 0, n + 1 < m ⊦ m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 - 1 = 0 ⊦ n0 + n = 0, n0 = 1, n0 = 0

Rule argument

Additional information

Used simplifier rules: lep: ⊦ m ≤ m + n;%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2