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Proof step 7 from proof minus-one-odd (previous step)

0 ≤ 2 * j0, 0 ≤ j0 ⊦ ~1 ^ (2 * j0) * ~1 = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 ≤ j0 ⊦ ~1 ^ (2 * j0) = 1

Rule argument

Additional information

Used simplifier rules: zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%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%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%minus-one-02: ⊦ ~1 * i = ~ i; from specification int-mult