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Proof step 7 from proof uint2bits-len-mon (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: uint2bits-one: ⊦ uint2bits(1) = false + true;%uint2bits-zero: ⊦ uint2bits(0) = false ';%uint2bits-zero: ⊦ uint2bits(0) = false ';%uint2bits-empty: 0 ≤ i ⊦ uint2bits(i) ≠ [];% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ 0 ≤ i ∧ i ≤ 1 → i = 0 ∨ i = 1; (forward)% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ boolvar + boolvar0 = boolvar ' + boolvar0 ';% ⊦ i ≤ j → (j ≤ i ↔ i = j);% ⊦ # boolvar ' = 0 +1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ # bits = 0 ↔ bits = [];% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ bits + [] = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar '.first ↔ boolvar;% ⊦ boolvar ' = boolvar0 ' ↔ boolvar ↔ boolvar0;