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Proof step 6 from proof dec-int-01 (previous step)

i < 0 ⊦ encode(2 * i→n(i) - 1) = i

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
2 * i→n(i) = m + 1 ∧ 2 * i→n(i) ≠ 0, i < 0 ⊦ encode(m) = i

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [2 * i→n(i), m] flags:

Additional information

Used simplifier rules: enc-zero-02: ⊦ encode(n) = 0 ↔ n = 0;%two-zero-01: ⊦ 2 * m = 0 ↔ m = 0; from specification nat-mult%abs-zero: ⊦ i→n(i) = 0 ↔ i = 0; from specification intnat%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: 2 * i→n(i), m