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Proof step 20 from proof uint2bits-falsein-01 (previous step)

2 ^ i = 2 * (i0 + 1), false ∈ uint2bits(i0).rest, 0 < i0, ∀ i. 2 ^ i ≠ i0 + 1, ∀ j0. i→n(j0) < i→n(i0 * 2 + 1) → 0 < j0 → (false ∈ uint2bits(j0).rest ↔ (∀ i. 2 ^ i ≠ j0 + 1)) ⊦ i0 = 0

User applied rule all left

And obtained the following new goals

Proof step 21:
2 ^ i = 2 * (i0 + 1), false ∈ uint2bits(i0).rest, 0 < i0, 2 ^ (i - 1) ≠ i0 + 1, ∀ i. 2 ^ i ≠ i0 + 1, ∀ j0. i→n(j0) < i→n(i0 * 2 + 1) → 0 < j0 → (false ∈ uint2bits(j0).rest ↔ (∀ i. 2 ^ i ≠ j0 + 1)) ⊦ i0 = 0

Rule argument

pos: left 4 terms: i - 1 abort: F discard: F computed: F

Additional information

Used terms: i - 1