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Proof step 4 from proof modex-odd-neg-01 (previous step)

⊦ (∃ j. i = 2 * j - 1 ∧ j ≤ 0) ↔ (∃ j. i = 2 * j + 1 ∧ j < 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
2 * j = i + 1, j ≤ 0 ⊦ ∃ j. i = 2 * j + 1 ∧ j < 0

Rule argument

Additional information

Used simplifier rules: forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs% ⊦ i ≤ ~1 ↔ i < 0; (s)% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; (s)% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * j + i = i * (j + 1); (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i - j = i0 ↔ i = i0 + j; (s)