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Proof step 10 from proof weakmon-lem (previous step)

m0 ≤ m1 + 1, ∀ m0, ii. m0 ≤ m1 ∧ (∀ n. ii(n) ≤ ii(n + 1)) ∧ (∀ m. ii(m) ≤ ii(m0)) → ii(m1) = ii(m0), ∀ n. ii(n) ≤ ii(n + 1), ∀ m. ii(m) ≤ ii(m0) ⊦ ii(m1 + 1) = ii(m0)

User applied rule all left

And obtained the following new goals

Proof step 11:
m0 ≤ m1 + 1, m0 ≤ m1 ∧ (∀ n. ii(n) ≤ ii(n + 1)) ∧ (∀ m. ii(m) ≤ ii(m0)) → ii(m1) = ii(m0), ∀ m. ii(m) ≤ ii(m0), ∀ n. ii(n) ≤ ii(n + 1), ∀ m0, ii. m0 ≤ m1 ∧ (∀ n. ii(n) ≤ ii(n + 1)) ∧ (∀ m. ii(m) ≤ ii(m0)) → ii(m1) = ii(m0) ⊦ ii(m1 + 1) = ii(m0)

Rule argument

pos: left 2 terms: m0, ii abort: F discard: F computed: T

Additional information

Used terms: m0, ii