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Proof step 10 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 = [] ⊦ bits2 ≠ [], bits0 = [], bitsand(bits + bits0, true ' + bits2) = bits + bitsand(bits0, true ' + bits2), # bits + # bits0 ≠ 1 ∧ bits2 ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ bits, bits0, bits1. # bits1 < # [] → # bits1 < # bits0 → bitsand(bits + bits0, true ' + bits1) = bits + bitsand(bits0, true ' + bits1) ∨ bits0 = [] ⊦ bits0 = [], bitsand(bits + bits0, true ') = bits + bits0

Rule argument

Additional information

Used simplifier rules: and-oner-01: ⊦ bits ≠ [] → bitsand(bits, true ') = bits;% ⊦ # bits = 0 ↔ bits = [];% ⊦ 0 < n ↔ n ≠ 0;% ⊦ # [] = 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + [] = bits;