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Proof step 4 from proof eq (previous step)

m1 +1 * m0 = n * m0, ∀ m0, n. m1 * m0 = n * m0 → m1 = n ∨ m0 = 0 ⊦ m1 +1 = n, m0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m0 * m1 + m0 = n * m0, ∀ m0, n. m1 * m0 = n * m0 → m1 = n ∨ m0 = 0 ⊦ m1 + 1 = n, m0 = 0

Rule argument

Additional information

Used simplifier rules: rsucc-02: ⊦ m * (n + 1) = m * n + m;%mult-a: ⊦ m0 * m * n = (m0 * m) * n;%mult-c: ⊦ m * n = n * m;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m +1 = m + 1; (s)