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Proof step 25 from proof band-max (previous step)

1 < #(false ' + true ') ∧ 1 < # int2bits(j), 1 < bits2int(bitsand((false ' + true ').butlast, int2bits(j).butlast) + ((false ' + true ').last ∧ int2bits(j).last)), 0 < bits2int(bitsand((false ' + true ').butlast, int2bits(j).butlast) + ((false ' + true ').last ∧ int2bits(j).last)), ∀ i0, j. i→n(i0) < i→n(1) → i0 < bits2int(bitsand(int2bits(i0), int2bits(j))) ∧ 0 ≤ i0 → false ⊦ j = ~1, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ bits + boolvar = bits + boolvar ';% ⊦ (bits + boolvar ').last ↔ boolvar;% ⊦ (bits + boolvar ').butlast = bits;% ⊦ bits + [] = bits;% ⊦ boolvar + bits = boolvar ' + bits;% ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ int2bits(j).first ↔ j < 0;% ⊦ # bits = 1 ↔ bits = bits.first + [];% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;