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Proof step 3 from proof num-i2n-eq-01 (previous step)

i + n→i(m) = 0, i < 0 ⊦ i→n(i) = m, m = 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = ~ n→i(m)

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);%n2n: ⊦ i→n(n→i(n)) = n;%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%neg-03: ⊦ ~ i + i = 0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 Used formulas: i = ~ n→i(m)