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Proof step 5 from proof eq (previous step)

m0 * m1 + m0 = n * m0, ∀ m0, n. m1 * m0 = n * m0 → m1 = n ∨ m0 = 0 ⊦ m1 + 1 = n, m0 = 0

User applied rule all left

And obtained the following new goals

Proof step 6:
m0 * m1 + m0 = n * m0, m1 * m0 = n -1 * m0 → m1 = n -1 ∨ m0 = 0, ∀ m0, n. m1 * m0 = n * m0 → m1 = n ∨ m0 = 0 ⊦ m1 + 1 = n, m0 = 0

Rule argument

pos: left 2 terms: m0, n -1 abort: F discard: F computed: F

Additional information

Used terms: m0, n -1