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Proof step 7 from proof aligndown-add-same0 (previous step)

n0 + n1 = m0 * m + n2, n0 = m1 * m + n3, n3 + n1 < m, n2 < m, n3 < m ⊦ (n2 = 0 ⊃ n0 + n1;m0 * m) = (n3 = 0 ⊃ n0;m1 * m), m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n3 + n1 < m, n3 < m ⊦ (n3 + n1 = 0 ⊃ (m0 * m + n3) + n1;m0 * m) = m0 * m

Rule argument

Additional information

Used simplifier rules: lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ m < n → n ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1); (ws)