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Proof step 29 from proof eq-one (previous step)

0 < i * j, 1 < i * j ⊦ i * j = 1 ↔ abs(j) = 1 ∧ i = j

User applied rule cut formula

And closed the goal!

Rule argument

fma: j < 0

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-gr-01: ⊦ i < 0 → (0 < i * j ↔ j < 0);%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%zero-gr-02: ⊦ 0 ≤ i → (0 < i * j ↔ 0 < j ∧ i ≠ 0);% ⊦ ¬ j < 0 → abs(j) = j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs Used formulas: j < 0