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Proof step 13 from proof int2bits-lslen-05 (previous step)

i < ~1, i < 0, # int2bits(i) ≤ m, 1 + (m - 1) < # int2bits(i), 1 + (m - 1) < m ⊦ m = 1

Heuristic var elimination applied rule insert elim lemma

And closed the goal!

Rule argument

seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m, m0] rotate: T spec: nat inst: name: elim-pred-c

Additional information

Used simplifier rules: int2bits-not-empty: ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m ≤ 0 ↔ m = 0;% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ m + 1 ≠ 0;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ ¬ n < n;%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0