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Proof step 8 from proof divide-list-at-value (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 9:
j < i, perm(ints0, ints1 + ints), ints = [] ∨ j < min(ints), ints1 = [] ∨ max(ints1) ≤ j ⊦ perm(i ' + ints0, ints1 + (i + ints)) ∧ (ints1 = [] ∨ max(ints1) ≤ j) ∧ (i + ints = [] ∨ j < min(i + ints))

Rule argument

pos: right 1 terms: ints1, i + ints abort: F discard: T computed: F

Additional information

Used terms: ints1, i + ints