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Proof step 4 from proof product-all-elems-le (previous step)

i0 ^ (1 + n→i(# ints0)) < i * Π ints0, ∀ i1. ¬ i1 ^ n→i(# ints0) < Π ints0 ∨ (∃ i. ¬(i ∈ ints0 → 0 < i ∧ i ≤ i1)), ∀ i1. i1 = i ∨ i1 ∈ ints0 → 0 < i1 ∧ i1 ≤ i0 ⊦

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 < 1 + n→i(# ints0), i0 ^ (1 + n→i(# ints0))-1 * i0 < i * Π ints0, ∀ i1. ¬ i1 ^ n→i(# ints0) < Π ints0 ∨ (∃ i. ¬(i ∈ ints0 → 0 < i ∧ i ≤ i1)), ∀ i1. i1 = i ∨ i1 ∈ ints0 → 0 < i1 ∧ i1 ≤ i0 ⊦

Rule argument

spec: int-pot inst: name: Pot-pos seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [1 + n→i(# ints0), i0] flags: rewrite

Additional information

Used simplifier rules: numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%nat2int: ⊦ 0 ≤ n→i(n); 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 Used terms: 1 + n→i(# ints0), i0