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Proof step 9 from proof rax-03 (previous step)

i0 + i1 + 1 = 0, i0 < i1, 0 ≤ i1, i0 + i1 < 0, i0 < 0 ⊦ i→n(i1) = i→n(i0 + 1)

User applied rule cut formula

And closed the goal!

Rule argument

fma: i1 = ~(i0 + 1)

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i);%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%neg-03: ⊦ ~ i + i = 0; from specification int-abs Used formulas: i1 = ~(i0 + 1)