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Proof step 1 from proof divide-list-at-value (conclusion)

⊦ ∃ ints0, ints1. perm(ints, ints0 + ints1) ∧ (ints0 = [] ∨ max(ints0) ≤ j) ∧ (ints1 = [] ∨ j < min(ints1))

User applied rule structural induction

And obtained the following new goals

Proof step 2:
⊦ ∃ ints0, ints1. perm([], ints0 + ints1) ∧ (ints0 = [] ∨ max(ints0) ≤ j) ∧ (ints1 = [] ∨ j < min(ints1))
Proof step 4:
∀ j. ∃ ints2, ints1. perm(ints0, ints2 + ints1) ∧ (ints2 = [] ∨ max(ints2) ≤ j) ∧ (ints1 = [] ∨ j < min(ints1)) ⊦ ∃ ints2, ints1. perm(i + ints0, ints2 + ints1) ∧ (ints2 = [] ∨ max(ints2) ≤ j) ∧ (ints1 = [] ∨ j < min(ints1))

Rule argument

indvar: ints

Additional information

Used simplifier rules: intlist freely generated by [], +;;