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Proof step 16 from proof sgn-lastn-01 (previous step)

¬ n < #(boolvar ' + boolvar ' + bits), n ≤ # sgn(boolvar ' + bits), ∀ n0. ¬ # sgn(boolvar ' + bits) < n0 → lastn(n0, sgn(boolvar ' + bits)) = lastn(n0, boolvar ' + bits) ⊦ lastn(n, sgn(boolvar ' + bits)) = lastn(n, boolvar ' + boolvar ' + bits)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
n ≤ # sgn(boolvar ' + bits), 2 + # bits ≤ n, 2 + # bits ≤ # sgn(boolvar ' + bits), ∀ n0. ¬ # sgn(boolvar ' + bits) < n0 → lastn(n0, sgn(boolvar ' + bits)) = lastn(n0, boolvar ' + bits) ⊦ lastn(n, sgn(boolvar ' + bits)) = lastn(n, boolvar ' + boolvar ' + bits)

Rule argument

Additional information

Used simplifier rules: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;