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Proof step 3 from proof app (previous step)

∀ a, y. a ∈ x0 → (x0 + y) -1l a = (x0 -1l a) + y, a ∈ a0 + x0 ⊦ ((a0 + x0) + y) -1l a = ((a0 + x0) -1l a) + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ x0, ∀ a, y. a ∈ x0 → (x0 + y) -1l a = (x0 -1l a) + y ⊦ a = a0, (x0 + y) -1l a = (x0 -1l a) + y

Rule argument

Additional information

Used simplifier rules: del1-y: ⊦ (a ' + x) -1l a = x;%del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a);%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%app-01: ⊦ x + y = x + z ↔ y = z; from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list