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Proof step 9 from proof subseteq-lebrefs-subseteq (previous step)

eq-entry(m, lref) ∈ q1, q ⊆ q1, ∀ q1. q ⊆ q1 → q.lebrefs ⊆ q1.lebrefs ⊦ lref = none, q.lebrefs ++ lref ⊆ q1.lebrefs

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 10:
eq-entry(m, lref) ∈ q1, q ⊆ q1, q ⊆ q1 → q.lebrefs ⊆ q1.lebrefs, ∀ q1. q ⊆ q1 → q.lebrefs ⊆ q1.lebrefs ⊦ lref = none, q.lebrefs ++ lref ⊆ q1.lebrefs

Rule argument

pos: left 3 terms: q1

Additional information

Used terms: q1