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Proof step 9 from proof bitsand-app-true (previous step)

# bits2 < # bits0, ∀ bits, bits0, bits1. # bits1 < # bits2 → # bits1 < # bits0 → bitsand(bits + bits0, true ' + bits1) = bits + bitsand(bits0, true ' + bits1) ∨ bits0 = [] ⊦ bits0 = [], bitsand(bits + bits0, true ' + bits2) = bits + bitsand(bits0, true ' + bits2), # bits + # bits0 ≠ 1 ∧ bits2 ≠ []

User applied rule cut formula

And obtained the following new goals

Proof step 10:
# bits2 < # bits0, ∀ bits, bits0, bits1. # bits1 < # bits2 → # bits1 < # bits0 → bitsand(bits + bits0, true ' + bits1) = bits + bitsand(bits0, true ' + bits1) ∨ bits0 = [] ⊦ bits2 ≠ [], bits0 = [], bitsand(bits + bits0, true ' + bits2) = bits + bitsand(bits0, true ' + bits2), # bits + # bits0 ≠ 1 ∧ bits2 ≠ []

Rule argument

fma: bits2 ≠ []

Additional information

Used simplifier rules: ⊦ # bits + # bits0 = 1 ∧ bits ≠ [] → # bits = 1 ∧ bits0 = []; (forward)% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ # [] = 0;% ⊦ 0 + m = m;% ⊦ [] + bits = bits;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1; Used formulas: bits2 ≠ []