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Proof step 4 from proof subapp (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ x0, y ⊆m y0, x1 ⊆m x0 -1l a, ∀ y, y0, x0. y ⊆m y0 ∧ x1 ⊆m x0 → 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;% ⊦ a ∈ x → (x + y) -1l a = (x -1l a) + y; (s)% ⊦ a ∈ x → a ∈ x + y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)