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Proof step 3 from proof uint2bits-len-mon (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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 ⊦

Proof step 3 from proof uint2bits-len-mon (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information