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Proof step 8 from proof uint2bits-falsein-01 (previous step)

uint2bits(j / 2) ≠ [], 0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (false ∈ uint2bits(j0).rest ↔ (∀ i. 2 ^ i ≠ j0 + 1)) ⊦ j = 1, false ∈ uint2bits(j / 2).rest + (j % 2 = 1)' ↔ (∀ i. 2 ^ i ≠ j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
0 < j, ∀ j0. i→n(j0) < i→n(j) → 0 < j0 → (false ∈ uint2bits(j0).rest ↔ (∀ i. 2 ^ i ≠ j0 + 1)) ⊦ j = 1, j % 2 ≠ 1 ∨ false ∈ uint2bits(j / 2).rest ↔ (∀ i. 2 ^ i ≠ j + 1)

Rule argument

Additional information

Used simplifier rules: uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ 0 ≤ i / 2 ↔ ~1 ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);