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Proof step 2 from proof num-i2n-eq-01 (previous step)

i < 0 ⊦ i→n(i) = m ↔ ~ i = n→i(m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i + n→i(m) = 0, i < 0 ⊦ i→n(i) = m, m = 0

Rule argument

Additional information

Used simplifier rules: i2i-neg: ⊦ i < 0 → n→i(i→n(i)) = ~ i;%num-n2i-ls-02: 0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1%ls-02: ⊦ i < i + j ↔ 0 < j; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%neg-12: ⊦ ~ i = ~ j ↔ i = j; from specification int-abs