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Proof step 31 from proof int2int-lastn-negneg (previous step)

# bits = # int2bits(i) - m, int2bits(i) = bits + lastn(m, int2bits(i)), i = (bits2int(bits) + 1) * 2 ^ n→i(m) + bits2int(lastn(m, int2bits(i))), 2 ^ n→i(m) + bits2int(lastn(m, int2bits(i))) ≤ 0, bits2int(lastn(m, int2bits(i))) < 2 ^ n→i(m), bits.first, m ≤ # int2bits(i), i < 0, 0 < 2 ^ n→i(m), i + 1 < 2 ^ n→i(m), i < 2 ^ n→i(m), bits2int(lastn(m, int2bits(i))) + 1 < 2 ^ n→i(m), bits2int(lastn(m, int2bits(i))) < 0 ⊦ i + 2 ^ n→i(m) = 0, bits2int(lastn(m, int2bits(i))) = bits2int(lastn(m, int2bits(i))) % 2 ^ n→i(m), m = 0, bits = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 32:
⊦ # bits ≤ n ∧ bits ≠ [] → ¬ 2 ^ n→i(n) + bits2int(bits) ≤ 0

Rule argument

name: bits2int-nle rotate: F subst: [bits, n] → [lastn(m, int2bits(i)), m]

Additional information

Used simplifier rules: ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ n ≤ # bits → # lastn(n, bits) = n;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ ¬ n < n;% ⊦ lastn(n, bits) = [] ↔ n = 0; Used terms: lastn(m, int2bits(i)), m