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Proof step 48 from proof equal-minus-one (previous step)

i ^ ~(j + 1) = ~ i, abs(~ i) = 1, i + ~ i = 0, j < 0 ⊦ i = 0, i = ~1, j = 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: i < 0

Additional information

Used simplifier rules: one-01: ⊦ 1 ^ j = 1;%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abs-neg: ⊦ abs(~ i) = abs(i); from specification int-abs%Abs-pos: ⊦ ¬ j < 0 → abs(j) = j; from specification int-abs%neg-03: ⊦ ~ i + i = 0; 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%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%numadd-01: ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; from specification int-abs Used formulas: i < 0