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Proof step 1 from proof div-mod (conclusion) (back to tree)

⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → (j | i - i0 ↔ i % j = i0 % j)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof div-mod (previous step) (back to tree)

¬(j | i - i0 ↔ i % j = i0 % j), 0 < j, 0 ≤ i, 0 ≤ i0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:div-mod-01, seq: ⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i → (j | i - i0 ↔ i % j = i0 % j), subst:[i0, i, j] → [i0, i, j], <all paths>

Additional information

Used terms: i0, i, j

Proof step 3 from proof div-mod (previous step) (back to tree)

⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i → (j | i - i0 ↔ i % j = i0 % j)

Lemma div-mod-01

Proof step 4 from proof div-mod (previous step) (back to tree)

¬(j | i - i0 ↔ i % j = i0 % j), 0 < j, 0 ≤ i0, 0 ≤ i ⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: i - i0 = ~(i0 - i)

Additional information

Used simplifier rules: ⊦ ~(i - j) = j - i; (s) ⊦ i - j = i0 ↔ i = i0 + j; (s) ⊦ i + (j - i) = j; (s) add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2 add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: i - i0 = ~(i0 - i)

Proof step 5 from proof div-mod (previous step) (back to tree)

i - i0 = ~(i0 - i), ¬(j | i - i0 ↔ i % j = i0 % j), 0 < j, 0 ≤ i, 0 ≤ i0 ⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i

User applied rule subst equation

And obtained the following new goals

Rule argument

pos: left 1, flags: discard

Additional information

Proof step 6 from proof div-mod (previous step) (back to tree)

¬(j | ~(i0 - i) ↔ i % j = i0 % j), 0 < j, 0 ≤ i0, 0 ≤ i ⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:div-neg, seq: ⊦ i | ~ j ↔ i | j, subst:[j, i] → [i0 - i, j], <all paths>

Additional information

Used terms: i0 - i, j

Proof step 7 from proof div-mod (previous step) (back to tree)

⊦ i | ~ j ↔ i | j

Lemma div-neg

Proof step 8 from proof div-mod (previous step) (back to tree)

¬(j | i0 - i ↔ i % j = i0 % j), 0 < j, 0 ≤ i, 0 ≤ i0 ⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:div-mod-01, seq: ⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i → (j | i - i0 ↔ i % j = i0 % j), subst:[i0, i, j] → [i, i0, j], <all paths>

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s) ⊦ i ≤ j ↔ ¬ j < i; (ss) ⊦ i < j → ¬ j < i; (s) Used terms: i, i0, j

Proof step 9 from proof div-mod (previous step) (back to tree)

⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i → (j | i - i0 ↔ i % j = i0 % j)

Lemma div-mod-01

Proof Tree for div-mod (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