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Proof step 63 from proof byte-law-and-one (previous step)

int2bits(i) ≠ [] ∧ int2bits(j) ≠ [], 8 = # fillsgn(8, int2bits(i)), # int2bits(i) < 7, # int2bits(i) < # int2bits(j), # int2bits(i) < # bitsand(int2bits(j), int2bits(i)), # int2bits(j) < 8, 8 < # bitsand(int2bits(j), int2bits(i)), # int2bits(j) + 1 < # bitsand(int2bits(j), int2bits(i)), # int2bits(j) < # bitsand(int2bits(j), int2bits(i)), # int2bits(i) + 1 < # bitsand(int2bits(j), int2bits(i)), # int2bits(i) < 8 ⊦ lastn(8, bitsand(int2bits(j), int2bits(i))) = lastn(8, fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j)))), 8 = # fillsgn(8, bitsand(int2bits(j), int2bits(i))), bitsand(fillsgn(8, int2bits(i)), int2bits(j)) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1;% ⊦ bitsand(int2bits(i), int2bits(i0)) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ int2bits(j) ≠ [];% ⊦ bits0 ≠ [] → bitsand(boolvar ' + bits, bits0) ≠ [];% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ int2bits(j).first ↔ j < 0;%⊦ bits ≠ [] ∧ # bits < n → fillsgn(n, bits) = bits.first ' + fillsgn(n -1, bits);