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Proof step 23 from proof bitsor-bi-zeroes (previous step)

¬(¬ (bits3 + bits1).first ∧ bits3 + bits1 ≠ []) ⊦ # bits3 = 1, bits2int(bits3 + bits1) = bits2uint(bits1), bits3 = [], true ∈ bits3

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
bits3.first ⊦ # bits3 = 1, bits2int(bits3 + bits1) = bits2uint(bits1), bits3 = [], true ∈ bits3

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ bits + bits0 ≠ [] ↔ bits = [] ∧ bits0 = [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits ≠ [] → ((bits + bits0).first ↔ bits.first);