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Proof step 5 from proof eq-pos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 7:
i0 ^ i -1 * i0 = i0 ^ j, 1 < i0, 0 < i, 2 ≤ i0, 0 ≤ i, 0 ≤ j, ∀ i1, i0, j. i→n(i1) < i→n(i) → i0 ^ i1 = i0 ^ j ∧ 1 < i0 ∧ 2 ≤ i0 ∧ 0 ≤ i1 ∧ 0 ≤ j → i1 = j ⊦ i = j

Rule argument

spec: <toplevel> name: Pot-pos seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [i, i0] flags: <all paths>

Additional information

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