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

y ⊆m y0, x ⊆m x0 ⊦ y + x ⊆m x0 + y0

User applied rule structural induction

And obtained the following new goals

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

Rule argument

indvar: y

Additional information

Used simplifier rules: subempty: ⊦ [] ⊆m x;%subapp-02: ⊦ x ⊆m x0 → x ⊆m x0 + y;%list freely generated by [], +;;% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list