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Proof step 3 from proof subapp (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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

Rule argument

Additional information

Used simplifier rules: subrec: ⊦ a ' + x ⊆m y ↔ ¬ ¬(a ∈ y ∧ x ⊆m y -1l a);%app: ⊦ a ∈ x → (x + y) -1l a = (x -1l a) + y; from specification list-del%in-11: ⊦ a ∈ x → a ∈ x + y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list