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Proof step 6 from proof uint2bits-2pot-05 (previous step)

uint2bits(i) = false ' + bits, 0 ≤ i, false ∈ uint2bits(i) ⊦ 2 ^ n→i(# bits) = i + 1, false ∈ bits

User applied rule cut formula

And obtained the following new goals

Proof step 7:
uint2bits(i) = false ' + bits, 0 ≤ i, false ∈ uint2bits(i) ⊦ bits = [], 2 ^ n→i(# bits) = i + 1, false ∈ bits

Rule argument

fma: bits = []

Additional information

Used simplifier rules: uint2bits-zero-01: 0 ≤ i ⊦ uint2bits(i) = false ' ↔ i = 0;% ⊦ bits + [] = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ boolvar ∈ [];% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0;% ⊦ # [] = 0;% ⊦ 0 < i ∧ 0 ≤ j → (1 = i + j ↔ i = 1 ∧ j = 0); Used formulas: bits = []