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Proof step 7 from proof and-size-one-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# bits2 = # bits1, ∀ bits, bits1, bits0. # bits < # bits2 → # bits = # bits1 → bits1 = [] ∨ bits0 = [] ∨ bitsand(true ' + bits, bits0 + bits1) = bits0 + bitsand(bits, bits1) ⊦ bits2 = [], bits1 = [], bits0 = [], bitsand(true ' + bits2.butlast, bits0 + bits1.butlast) + (bits2.last ∧ bits1.last)' = bits0 + bitsand(bits2, bits1), # bits0 + # bits1 = 0

Rule argument

Additional information

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