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Proof step 127 from proof byte-law-and-one (previous step)

8 = # fillsgn(8, bitsand(bits, int2bits(i))), 7 = # fillsgn(7, int2bits(i)), int2bits(j) = bits0 + bits, # bits = 8, # int2bits(i) < # bitsand(bits, false ' + fillsgn(7, int2bits(i))), 0 ≤ i, # int2bits(i) < 8, 8 < # bitsand(bits, false ' + fillsgn(7, int2bits(i))), # int2bits(i) + 1 < # bitsand(bits, false ' + fillsgn(7, int2bits(i))) ⊦ fillsgn(8, bitsand(bits, int2bits(i))) = lastn(8, bitsand(bits, false ' + fillsgn(7, int2bits(i)))), bits0 + bits = [], 8 = # fillsgn(8, bitsand(bits, false ' + fillsgn(7, int2bits(i)))), bitsand(bits, false ' + fillsgn(7, int2bits(i))) = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ ¬ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits0 subst: [bits0, bits] → [false ' + fillsgn(7, int2bits(i)), bits] rotate: F spec: intops inst: name: bitsand-len-false

Additional information

Used simplifier rules: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ boolvar ' ≠ [];% ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ boolvar ' + bits ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ # boolvar ' = 0 +1;% ⊦ m + 1 ≤ n ↔ m < n;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ ¬ # fillsgn(n, bits) < # bits;%n0 < n ⊦ (* n) < m0 + (* n0) ↔ n - n0 < m0;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ (boolvar ' + bits).first ↔ boolvar;% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward) Used terms: false ' + fillsgn(7, int2bits(i)), bits