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Proof step 1 from proof num-i2n-eq-01 (conclusion) (back to tree)

i < 0 ⊦ i→n(i) = (* m) ↔ ~ i = n→i(m)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

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

¬(i→n(i) = m ↔ ~ i = n→i(m)), i < 0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: i2i-neg: ⊦ i < 0 → n→i(i→n(i)) = ~ i; ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs ⊦ ~ i = j ↔ i + j = 0; (s) ⊦ ~ i = ~ j ↔ i = j; (s)

Proof step 3 from proof num-i2n-eq-01 (previous step) (back to tree)

i + n→i(m) = 0, i→n(i) ≠ m, i < 0 ⊦

User applied rule cut formula

And closed the goal!

Rule argument

fma: i = ~ n→i(m)

Additional information

Used simplifier rules: abs-neg: ⊦ i→n(~ i) = i→n(i); n2n: ⊦ i→n(n→i(n)) = n; ⊦ ~ i = j ↔ i + j = 0; (s) add-c: ⊦ i + j = j + i; from specification int-basic2 ⊦ ~ i + i = 0; (s) add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 Used formulas: i = ~ n→i(m)

Proof Tree for num-i2n-eq-01 (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree