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Proof step 10 from proof eq-pos (previous step)

i0 ^ i2 * i0 = i0 ^ j, ~2 < i2, 1 < i0, 0 ≤ j, 2 ≤ i0, 0 ≤ i2, ~1 ≤ i2, ∀ i1, i0, j. i→n(i1) < i→n(i2 +1) → i0 ^ i1 = i0 ^ j ∧ 1 < i0 ∧ 2 ≤ i0 ∧ 0 ≤ i1 ∧ 0 ≤ j → i1 = j ⊦ i2 + 1 = j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 12:
i0 ^ i2 * i0 = i0 ^ j -1 * i0, 1 < i0, ~2 < i2, 0 < j, ~1 ≤ i2, 0 ≤ i2, 2 ≤ i0, 0 ≤ j, ∀ i1, i0, j. i→n(i1) < i→n(i2 +1) → i0 ^ i1 = i0 ^ j ∧ 1 < i0 ∧ 2 ≤ i0 ∧ 0 ≤ i1 ∧ 0 ≤ j → i1 = j ⊦ i2 + 1 = j

Rule argument

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

Additional information

Used simplifier rules: Pot-zero: ⊦ j ^ 0 = 1;% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)% ⊦ i * j = 1 ↔ abs(i) = 1 ∧ i = j; (ws)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ (* j) = i0 + (* j0) ↔ j - j0 = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)% ⊦ ~1 < i ↔ 0 ≤ i; (s)%lf-03: ⊦ i < j ∧ i = i0 → i0 < j; (forward) from specification int-basic1%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1% ⊦ 1 < i → abs(i) = i; (s)% ⊦ ¬ i < i; (s) Used terms: j, i0