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Proof step 7 from proof dec-int (previous step)

2 * i→n(i) = m + 1 ∧ 2 * i→n(i) ≠ 0, i < 0 ⊦ decode(m) = i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
2 * i→n(i) = m + 1, i < 0 ⊦ decode(m) = i, i = 0

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%abs-zero: ⊦ i→n(i) = 0 ↔ i = 0; from specification intnat%two-zero-01: ⊦ 2 * m = 0 ↔ m = 0; from specification nat-mult