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Proof step 7 from proof even-not-ex (previous step)

i = i1 * 2 + j0, i < 0, abs(j0) < 2, 0 < 2, j0 ≤ 0, i1 ≤ 0 ⊦ j0 = 0 ↔ ¬(∃ i0. i = 2 * i0 + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: nummult-ls: ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); (s)% ⊦ i * j + i = i * (j + 1); (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i < 1 ↔ i ≤ 0; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)%i < 0 ⊦ (* i) + j = j - ~ i; (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)%forward-le-04: ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i < 0 → (~2 < i ↔ i = ~1); (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ 2 * i ≠ 2 * 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 + 0 = i; (s)