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Proof step 25 from proof and-first (previous step)

1 < #(boolvar ' + bits1) ∧ 1 < #(boolvar0 ' + bits), bitsand((boolvar ' + bits1).butlast, (boolvar0 ' + bits).butlast) + ((boolvar ' + bits1).last ∧ (boolvar0 ' + bits).last) = boolvar1 ' + bits0, ∀ bits2, boolvar, bits1, boolvar0, boolvar1, bits0. # bits2 < # bits → bitsand(boolvar ' + bits1, boolvar0 ' + bits2) = boolvar1 ' + bits0 → (boolvar1 ↔ boolvar ∧ boolvar0) ⊦ boolvar1 ↔ boolvar ∧ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
bitsand(boolvar ' + bits1.butlast, boolvar0 ' + bits.butlast) + (bits1.last ∧ bits.last)' = boolvar1 ' + bits0, ∀ bits2, boolvar, bits1, boolvar0, boolvar1, bits0. # bits2 < # bits → bitsand(boolvar ' + bits1, boolvar0 ' + bits2) = boolvar1 ' + bits0 → (boolvar1 ↔ boolvar ∧ boolvar0) ⊦ bits = [], bits1 = [], boolvar1 ↔ boolvar ∧ boolvar0, ¬ (bits1.last ∧ bits.last) ∈ bits0 ∧ ¬(bits1.last ∧ bits.last ↔ boolvar1), ¬(boolvar1 ↔ bits1.last ∧ bits.last) ∧ ¬ boolvar1 ∈ bitsand(boolvar ' + bits1.butlast, boolvar0 ' + bits.butlast)

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ bits + boolvar ' = bits0 → boolvar ∈ bits0; (forward)% ⊦ bits ≠ [] → ((boolvar ' + bits).last ↔ bits.last);% ⊦ bits ≠ [] → (bits0 + bits).butlast = bits0 + bits.butlast;% ⊦ # 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);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + boolvar = bits + boolvar ';