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Proof step 2 from proof one-03 (previous step)

1 = i + j, 0 < i ⊦ j < 0, i = 1 ∧ j = 0

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = 1 - j

Additional information

Used simplifier rules: sub-add-01: ⊦ i + (j - i0) = i + j - i0;%sub-add: ⊦ i + j - i = j;%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j;%lsone: ⊦ i < 1 ↔ i ≤ 0;%le-def: ⊦ i ≤ j ↔ ¬ j < i;%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%lst-02: ⊦ ¬ i < j → (¬ j < i ↔ i = j); from specification int-basic1 Used formulas: i = 1 - j