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Proof step 15 from proof put-in-01 (
previous step
)
m0 < # x0, ∀ m, a, b. m < # x0 → (a ∈ x0[m, b] ↔ a = b ∨ a ∈ x0 -1l x0[m]) ⊦ a ∈ x0[m0, b] ↔ a = b ∨ a ∈ x0 -1l x0[m0], x0[m0] = a0, a = a0, x0 = []
User applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: m0, a, b abort: F discard: F computed: T
Additional information
Used terms: m0, a, b