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Proof step 3 from proof modmod-add-if3 (previous step)

0 < j, 0 < j ⊦ (0 ≤ i ⊃ (0 ≤ i0 ⊃ (i + i0) % j;(0 ≤ i + i0 ⊃ (i + i0) % j;(i + i0 % j < 0 ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j + j;0)))) ;(i0 < 0 ⊃ (i + i0) % j;(i + i0 < 0 ⊃ (i + i0) % j;(0 ≤ i + i0 % j ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j - j;0)))) ) = ( 0 ≤ i + i0 % j ∧ i + i0 < 0 ∧ 0 ≤ i ∧ i0 < 0 ∧ (0 ≤ i ⊃ (0 ≤ i0 ⊃ (i + i0) % j;(0 ≤ i + i0 ⊃ (i + i0) % j;(i + i0 % j < 0 ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j + j;0)))) ;(i0 < 0 ⊃ (i + i0) % j;(i + i0 < 0 ⊃ (i + i0) % j;(0 ≤ i + i0 % j ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j - j;0)))) ) ≠ 0 ⊃ (i + i0) % j + j ;( i + i0 % j < 0 ∧ 0 ≤ i + i0 ∧ i < 0 ∧ 0 ≤ i0 ∧ (0 ≤ i ⊃ (0 ≤ i0 ⊃ (i + i0) % j;(0 ≤ i + i0 ⊃ (i + i0) % j;(i + i0 % j < 0 ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j + j;0)))) ;(i0 < 0 ⊃ (i + i0) % j;(i + i0 < 0 ⊃ (i + i0) % j;(0 ≤ i + i0 % j ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j - j;0)))) ) ≠ 0 ⊃ (i + i0) % j - j ;(i + i0) % j ) )

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-neq-01: j ≠ 0 ⊦ i % j + j ≠ 0;%mod-neq: j ≠ 0 ⊦ i % j ≠ j;%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs