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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: del1-n: ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a);%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%in-06: ⊦ ¬ 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