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Proof step 3 from proof subapp-01 (previous step)

∀ x, x0, y0. x ⊆m x0 ∧ y1 ⊆m y0 → y1 + x ⊆m x0 + y0, x ⊆m x0, a + y1 ⊆m y0 ⊦ (a + y1) + x ⊆m x0 + y0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
x ⊆m x0, a ∈ y0, y1 ⊆m y0 -1l a, ∀ x, x0, y0. x ⊆m x0 ∧ y1 ⊆m y0 → y1 + x ⊆m x0 + y0 ⊦ y1 + x ⊆m (x0 + y0) -1l a

Rule argument

Additional information

Used simplifier rules: subrec: ⊦ a ' + x ⊆m y ↔ ¬ ¬(a ∈ y ∧ x ⊆m y -1l a);%in-12: ⊦ a ∈ y → a ∈ x + y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list