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Proof step 6 from proof divide-list-at-value (previous step)

∃ ints2, ints1. perm(ints0, ints2 + ints1) ∧ (ints2 = [] ∨ max(ints2) ≤ j) ∧ (ints1 = [] ∨ j < min(ints1)) ⊦ ∃ ints, ints1. perm(i ' + ints0, ints + ints1) ∧ (ints = [] ∨ max(ints) ≤ j) ∧ (ints1 = [] ∨ j < min(ints1))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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))

Proof step 6 from proof divide-list-at-value (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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