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Proof step 1 from proof modex-odd (conclusion) (back to tree)

⊦ i % 2 = 1 ↔ (∃ j. i = 2 * j + 1 ∧ 0 ≤ j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-two-04: ⊦ 0 ≤ i → (i * 2 + 1) % 2 = 1; add-c: ⊦ i + j = j + i; from specification int-basic2 add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 commutativity: ⊦ i * j = j * i; from specification int-mult associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult

Proof step 2 from proof modex-odd (previous step) (back to tree)

i % 2 = 1 ⊦ ∃ j. i = 2 * j + 1 ∧ 0 ≤ j

User applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:div-elim-pospos, seq:⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0), subst:[i, j, i0, j0] → [i, 2, i0, j0]

Additional information

Used simplifier rules: nummult-eq-01: j ≠ 0, j0 % j = 0 ⊦ i * (* j) = i0 * (* j0) ↔ i = i0 * j0 / j; ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); (s) ⊦ i + j = i + i0 ↔ j = i0; (s) ⊦ 1 * i = i; (s) associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult commutativity: ⊦ i * j = j * i; from specification int-mult add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) Used terms: i, 2, i0, j0

Proof step 3 from proof modex-odd (previous step) (back to tree)

⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)

Lemma div-elim-pospos

Proof step 4 from proof modex-odd (previous step) (back to tree)

i % 2 = 1 ⊦ ∃ j. i = 2 * j + 1 ∧ 0 ≤ j, 0 < 2 ∧ 0 ≤ i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs ⊦ i ≤ j ↔ ¬ j < i; (ss)

Proof step 5 from proof modex-odd (previous step) (back to tree)

i % 2 = 1, i < 0 ⊦ ∃ j. i = 2 * j + 1 ∧ 0 ≤ j

User applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:div-elim-negpos, seq:⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0), subst:[i, j, i0, j0] → [i, 2, i0, j0]

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) Used terms: i, 2, i0, j0

Proof step 6 from proof modex-odd (previous step) (back to tree)

⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)

Lemma div-elim-negpos

Proof Tree for modex-odd (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