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Proof step 5 from proof minus-odd (previous step)

0 < j, ∃ j0. j = j0 + j0 + 1 ⊦ i ^ j + ~ i ^ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 ≤ j0 ⊦ i ^ (2 * j0 + 1) + ~ i ^ (2 * j0 + 1) = 0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + i = 2 * i; (s)