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Proof step 3 from proof subapp (previous step)

y ⊆m y0, x ⊆m x0 ⊦ x + y ⊆m x0 + y0

User applied rule structural induction

And obtained the following new goals

Proof step 4:
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

Rule argument

indvar: x

Additional information

Used simplifier rules: subempty: ⊦ [] ⊆m x;%subapp-03: ⊦ x ⊆m x0 → x ⊆m y + x0;%list freely generated by [], +;;% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list