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Proof step 53 from proof bitsand-cast-01 (previous step)

¬(# bits1 ≤ # bits ∧ bits1 ≠ [] ∧ ¬ bits1.first), # bits1 ≤ # bitsand(bits3 + bits, bits1), # bits = # bits1, # bits1 = # bitsand(bits1, bits) ⊦ lastn(# bits1, bitsand(bits3 + bits, bits1)) = bitsand(bits1, bits), bits3 = [], bits1 = [], bits3 + bits = [], bits2 = [], bits2 + bits1 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 54:
# bits1 = # bitsand(bits1, bits), # bits = # bits1, bits1.first ⊦ lastn(# bits1, bits3 + bitsand(bits, bits1)) = bitsand(bits1, bits), bits2 + bits1 = [], bits2 = [], bits3 + bits = [], bits1 = [], bits3 = []

Rule argument

Additional information

Used simplifier rules: bitsand-app-true: ⊦ # bits1 ≤ # bits0 ∧ bits1 ≠ [] ∧ bits1.first → bitsand(bits + bits0, bits1) = bits + bitsand(bits0, bits1);%bitsand-len-true: ⊦ bits0.first ∧ # bits0 ≤ # bits ∧ bits0 ≠ [] → # bitsand(bits, bits0) = # bits;% ⊦ m ≤ m;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m ≤ m + n;% ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ ¬ n < n;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;