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Proof step 3 from proof minus-one (previous step)

n→i(m) = i + 1, i < 0 ⊦ m = 0 ∧ i = ~1

User applied rule cut formula

And obtained the following new goals

Proof step 4:
n→i(m) = i + 1, i < 0, i + 1 < 0 ⊦ m = 0 ∧ i = ~1

Rule argument

fma: i + 1 < 0

Additional information

Used simplifier rules: nat2int-02: ⊦ n→i(n) = 0 ↔ n = 0;% ⊦ i + (* j) < (* j0) ↔ i < j0 - j; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < 0 → (~1 ≤ i ↔ i = ~1); (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lf-03: ⊦ i < j ∧ i = i0 → i0 < j; (forward) from specification int-basic1 Used formulas: i + 1 < 0