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Proof step 10 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]) ⊦ a ∈ (a0 ' + x0)[m, b] ↔ a = b ∨ a ∈ (a0 ' + x0) -1l (a0 ' + x0)[m], (a0 ' + x0)[m] = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m ≤ # x0, ∀ m, a, b. m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0 -1l x0[m]) ⊦ a ∈ (a0 ' + x0)[m, b] ↔ a = b ∨ a ∈ a0 ' + (x0 -1l (a0 ' + x0)[m]), (a0 ' + x0)[m] = a0

Rule argument

Additional information

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