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Proof step 4 from proof eq-pos (previous step)

i0 ^ i = i0 ^ j, 2 ≤ i0, 0 ≤ j, 0 ≤ i, 1 < i0, ∀ i1, i0, j. i→n(i1) < i→n(i) → i0 ^ i1 = i0 ^ j ∧ 2 ≤ i0 ∧ 0 ≤ j ∧ 0 ≤ i1 ∧ 1 < i0 → i1 = j ⊦ i = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 6:
0 < i, i0 ^ i -1 * i0 = i0 ^ j, 2 ≤ i0, 0 ≤ j, 0 ≤ i, 1 < i0, ∀ i1, i0, j. i→n(i1) < i→n(i) → i0 ^ i1 = i0 ^ j ∧ 2 ≤ i0 ∧ 0 ≤ j ∧ 0 ≤ i1 ∧ 1 < i0 → i1 = j ⊦ i = j

Rule argument

spec: int-pot inst: name: Pot-pos seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [i, i0] flags: rewrite, current

Additional information

Used simplifier rules: Pot-zero: ⊦ j ^ 0 = 1;%eqone-01: ⊦ i ^ j = 1 ∧ 0 < j → abs(i) = 1;%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs%lf-03: ⊦ i < j ∧ i = i0 → i0 < j; (forward) from specification int-basic1%abs: ⊦ 1 < i → abs(i) = i; from specification int-abs%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1 Used terms: i, i0