[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 14 from proof mod-pot-shift-pos (previous step)

0 < j, 0 ≤ i2, ∀ i1, j, i. i→n(i1) < i→n(i2 + 1) → 0 < j ∧ 0 < i1 → i1 = 0 ∨ (i % j) ^ i1 % j = i ^ i1 % j ⊦ 0 = j, ~1 = i2, ((i % j) ^ i2 % j * i % j) % j = (i ^ i2 * i) % j

User applied rule apply induction

And obtained the following new goals

Proof step 15:
0 < j, 0 ≤ i2, ∀ i1, j, i. i→n(i1) < i→n(i2 + 1) → 0 < j ∧ 0 < i1 → i1 = 0 ∨ (i % j) ^ i1 % j = i ^ i1 % j ⊦ 0 < j ∧ 0 < i2, 0 = j, ~1 = i2, ((i % j) ^ i2 % j * i % j) % j = (i ^ i2 * i) % j
Proof step 18:
i2 = 0 ∨ (i % j) ^ i2 % j = i ^ i2 % j, 0 < j, 0 ≤ i2, ∀ i1, j, i. i→n(i1) < i→n(i2 + 1) → 0 < j ∧ 0 < i1 → i1 = 0 ∨ (i % j) ^ i1 % j = i ^ i1 % j ⊦ 0 = j, ~1 = i2, ((i % j) ^ i2 % j * i % j) % j = (i ^ i2 * i) % j

Rule argument

subst: [i1, j, i] → [i2, j, i]

Additional information

Used simplifier rules: rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1; from specification intnat%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%irref: ⊦ ¬ i < i; from specification int-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: i2, j, i