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Proof step 16 from proof dec-int-01 (previous step)

m0 + 1 = i→n(i), i < 0 ⊦ i = 0, ~1 = n→i(m0) + i

User applied rule cut formula

And obtained the following new goals

Proof step 17:
m0 + 1 = i→n(i), i < 0 ⊦ n→i(m0 + 1) = n→i(i→n(i)), i = 0, ~1 = n→i(m0) + i

Rule argument

fma: n→i(m0 + 1) = n→i(i→n(i))

Additional information

Used simplifier rules: succ-01: ⊦ n→i(m + 1) = n→i(m) + 1; from specification intnat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%i2i-neg: ⊦ i < 0 → n→i(i→n(i)) = ~ i; from specification intnat%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%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 Used formulas: n→i(m0 + 1) = n→i(i→n(i))