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Proof step 3 from proof bitsxor-rec-01 (previous step)

bits = bits1 + boolvar ' ⊦ bitsxor(bits, bits0) = bitsxor(bits1, bits0.butlast) + (¬(boolvar ↔ bits0.last))', # bits = 1, bits = [], # bits0 = 1, bits0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ # bits0 = 1, bits0 = [], bitsxor(bits1 + boolvar ', bits0) = bitsxor(bits1, bits0.butlast) + (¬(boolvar ↔ bits0.last))', bits1 = []

Rule argument

Additional information

Used simplifier rules: ⊦ bits + boolvar ' ≠ [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;