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Proof step 6 from proof uint2bits-len-mon (previous step)

¬(0 < j ∧ j ≠ 1), 0 ≤ i, i ≤ j, # uint2bits(j) < # uint2bits(i), 0 ≤ j, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ j0 ∧ # uint2bits(j0) < # uint2bits(i) ∧ i ≤ j0 ∧ 0 ≤ i → false ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
0 ≤ i, i ≤ j, # uint2bits(j) < # uint2bits(i), 0 ≤ j, ∀ j0, i. i→n(j0) < i→n(j) → 0 ≤ j0 ∧ # uint2bits(j0) < # uint2bits(i) ∧ i ≤ j0 ∧ 0 ≤ i → false ⊦ j ≠ 1 ∧ 0 ≠ j

Rule argument

Additional information

Used simplifier rules: uint2bits-zero: ⊦ uint2bits(0) = false ';%uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # boolvar ' = 0 +1;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ ¬ i < j ↔ j ≤ i;