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Proof step 8 from proof aligndown-add-same0 (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m = m + n ↔ n = 0; (s)