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Proof step 3 from proof n2i-mult (previous step)

∀ m. n→i(m * n0) = n→i(m) * n→i(n0) ⊦ n→i(m * n0) + n→i(m) = n→i(m) * (n→i(n0) + 1)

User applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: m abort: F discard: F computed: F

Additional information

Used simplifier rules: inj: ⊦ n→i(m) = n→i(n) ↔ m = n;%eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0; from specification int-mult%one: ⊦ 1 * i = i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: m