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Proof step 6 from proof cast-if (previous step)

j < 0 ∨ 2 ^ (i - 1) ≤ j % 2 ^ i, 0 < i ⊦ cast(i, j) = (¬ j < 0 ∧ ¬ j % 2 ^ i < 2 ^ (i - 1) ⊃ j % 2 ^ i - 2 ^ i ;(¬ 0 ≤ j ∧ ¬ 0 ≤ j % 2 ^ i + 2 ^ (i - 1) ⊃ j % 2 ^ i + 2 ^ i;(¬ 0 ≤ j ∧ ¬ 2 ^ (i - 1) + j % 2 ^ i < 0 ⊃ j % 2 ^ i;j)) )

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
¬ j < 0 ∧ ¬ j % 2 ^ i < 2 ^ (i - 1), j < 0 ∨ 2 ^ (i - 1) ≤ j % 2 ^ i, 0 < i ⊦ cast(i, j) = j % 2 ^ i - 2 ^ i
Proof step 10:
j < 0 ∨ j % 2 ^ i < 2 ^ (i - 1), j < 0 ∨ 2 ^ (i - 1) ≤ j % 2 ^ i, 0 < i ⊦ cast(i, j) = (¬ 0 ≤ j ∧ ¬ 0 ≤ j % 2 ^ i + 2 ^ (i - 1) ⊃ j % 2 ^ i + 2 ^ i;(¬ 0 ≤ j ∧ ¬ 2 ^ (i - 1) + j % 2 ^ i < 0 ⊃ j % 2 ^ i;j))

Rule argument

pos: right 1

Proof step 6 from proof cast-if (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information