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Proof step 634 from proof bitsand-cast (previous step)

2 + # bits + m ≤ #((bits2 + mklist(false, m)) + bitsand(false ' + true ' + bits, bits1)), 2 + # bits = # bits1, # bits1 = # bitsand(bits1, false ' + true ' + bits), bits3.first, # bits3 ≤ m ⊦ bits3 + bits1 = [], bits2 + mklist(false, m) = [], bits1 = [], mklist(false, m) + bitsand(bits1, false ' + true ' + bits) = lastn(2 + # bits + m, (bits2 + mklist(false, m)) + bitsand(false ' + true ' + bits, bits1)), bits3 = [], m = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 635:
2 + # bits = # bits1, # bits1 = # bitsand(bits1, false ' + true ' + bits), bits3.first, # bits3 ≤ m, # bits1 ≤ # bits2 + # bitsand(false ' + true ' + bits, bits1) ⊦ bits3 + bits1 = [], bits2 + mklist(false, m) = [], bits1 = [], mklist(false, m) + bitsand(bits1, false ' + true ' + bits) = lastn(2 + # bits + m, (bits2 + mklist(false, m)) + bitsand(false ' + true ' + bits, bits1)), bits3 = [], m = 0

Rule argument

Additional information

Used simplifier rules: ⊦ m + n ≤ m + n0 ↔ n ≤ n0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # mklist(boolvar, n) = n;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;