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Proof step 10 from proof and-false (previous step)

¬(1 < #(true ' + bits2) ∧ 1 < #(true ' + bits)), bitsand(true ' + bits2, true ' + bits) = false ' + bits1, ∀ bits1, bits, bits0. # bits0 < # bits2 → bitsand(true ' + bits0, true ' + bits) = false ' + bits1 → false ⊦

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: and-onel-02: ⊦ bitsand(true ', boolvar ' + bits) = boolvar ' + bits;%and-oner: ⊦ bitsand(boolvar ' + bits, true ') = boolvar ' + bits;%⊦ boolvar ' + bits = boolvar0 ' + bits0 ↔ (boolvar ↔ boolvar0) ∧ bits = bits0;% ⊦ bits + [] = bits;% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 = m + 1 ↔ m = 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;