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Proof step 190 from proof byte-law-and-one (previous step)

# int2bits(i) ≤ # bits1 ∧ int2bits(i) ≠ [] ∧ int2bits(i).first, # bits2 + # bits1 = 8, int2bits(j) = (bits0 + bits2) + bits1, # bits1 = # int2bits(i), 7 = # fillsgn(7, int2bits(i)), 8 ≤ # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), 8 ≤ # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < 8, i < 0, # bits1 < # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) ⊦ true ' + fillsgn(7, int2bits(i)) = fillsgn(8, int2bits(i)), bitsand(bits2 + bits1, true ' + fillsgn(7, int2bits(i))) = [], lastn(8, fillsgn(8, (bits0 + bits2) + bitsand(bits1, int2bits(i)))) = lastn(8, bitsand(bits2 + bits1, true ' + fillsgn(7, int2bits(i)))), bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) = [], bits2 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ bits ≠ [] ∧ # bits < n → fillsgn(n, bits) = bits.first ' + fillsgn(n -1, bits) subst: [bits, n] → [int2bits(i), 8] rotate: F spec: bitopsaux inst: name: fill-rec

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + bits0 = bits + bits1 ↔ bits0 = bits1;% ⊦ int2bits(j) ≠ [];% ⊦ j < 0 → int2bits(j).first;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ n ≤ # bits → fillsgn(n, bits) = bits;% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m + n = n + m;% ⊦ m ≤ n ∧ n < n0 → m < n0; (forward)%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) Used terms: int2bits(i), 8