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Proof step 5 from proof int2int-lastn-posneg (previous step)

# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), m ≤ # int2bits(i), lastn(m, int2bits(i)).first, 0 < i ⊦ i % 2 ^ n→i(m) = bits2int(lastn(m, int2bits(i))) + 2 ^ n→i(m), m = 0

User applied rule cut formula

And obtained the following new goals

Proof step 6:
# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), m ≤ # int2bits(i), lastn(m, int2bits(i)).first, 0 < i ⊦ bits = [], i % 2 ^ n→i(m) = bits2int(lastn(m, int2bits(i))) + 2 ^ n→i(m), m = 0

Rule argument

fma: bits = []

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];%int2int-inj: ⊦ bits2int(int2bits(i)) = i;%int2bits-first-02: ⊦ int2bits(j).first ↔ j < 0;% ⊦ # [] = 0;% ⊦ m ≤ n → (n - m = 0 ↔ n = m);% ⊦ [] + bits = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ i < j → ¬ j < i; Used formulas: bits = []