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Proof step 9 from proof put-in-01 (previous step)

m ≤ # x0, ∀ m, a, b. m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0 -1l x0[m]) ⊦ (a0 ' + x0)[m] = a0, a ∈ (a0 ' + x0)[m, b] ↔ a = b ∨ a ∈ (a0 ' + x0) -1l (a0 ' + x0)[m]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
m ≤ # x0, ∀ m, a, b. m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0 -1l x0[m]) ⊦ (a0 ' + x0)[m] = a0, a ∈ (a0 ' + x0)[m, b] ↔ a = b ∨ a = a0 ∨ a ∈ x0 -1l (a0 ' + x0)[m]

Rule argument

Additional information

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