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Proof step 5 from proof subapp-01 (previous step)

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

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: y0 -1l a, x, x0 abort: F discard: T computed: T

Additional information

Used simplifier rules: subdel-01: ⊦ x ⊆m y + (y0 -1l a) ∧ a ∈ y0 → x ⊆m (y + y0) -1l a;%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: y0 -1l a, x, x0