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Proof step 4 from proof subapp (previous step)

y ⊆m y0, a ∈ x0, x1 ⊆m x0 -1l a, ∀ x0, y, y0. x1 ⊆m x0 ∧ y ⊆m y0 → x1 + y ⊆m x0 + y0 ⊦ x1 + y ⊆m (x0 -1l a) + y0

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 5:
y ⊆m y0, a ∈ x0, x1 ⊆m x0 -1l a, x1 ⊆m y ∧ y0 ⊆m x0 -1l a → x1 + y0 ⊆m y + (x0 -1l a), ∀ x0, y, y0. x1 ⊆m x0 ∧ y ⊆m y0 → x1 + y ⊆m x0 + y0 ⊦ x1 + y ⊆m (x0 -1l a) + y0

Rule argument

pos: left 4 terms: y, y0, x0 -1l a

Additional information

Used terms: y, y0, x0 -1l a