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Proof step 3 from proof div-add-succ (previous step)

0 < i, 0 < j, 1 < j ⊦ i % j = 0, (i + j - 1) / j = i / j + 1

User applied rule cut formula

And obtained the following new goals

Proof step 4:
i + j - 1 = (i - 1) + j, 1 < j, 0 < j, 0 < i ⊦ i % j = 0, (i + j - 1) / j = i / j + 1

Rule argument

fma: i + j - 1 = (i - 1) + j

Additional information

Used simplifier rules: ⊦ i + (j - i0) = i + j - i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + (j - i) = j; (s)% ⊦ i + j = i + i0 ↔ j = i0; (s) Used formulas: i + j - 1 = (i - 1) + j