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Proof step 4 from proof fill-one-02 (previous step)

# boolvar ' ≤ i→n(i) ∧ boolvar ' ≠ [], 0 < i ⊦ mklist(boolvar '.first, i→n(i) - # boolvar ') + boolvar ' = mklist(boolvar, i→n(i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
0 < i ⊦ boolvar ' + mklist(boolvar, i→n(i) - 1) = mklist(boolvar, i→n(i))

Rule argument

Additional information

Used simplifier rules: ⊦ 1 ≤ i ↔ 0 < i;%0 ≤ i ⊦ (* m) ≤ i→n(i) ↔ n→i(m) ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j → ¬ j < i;% ⊦ boolvar ' ≠ [];% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ mklist(boolvar, n) + boolvar ' = boolvar ' + mklist(boolvar, n);% ⊦ # boolvar ' = 0 +1;% ⊦ boolvar '.first ↔ boolvar;