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Proof step 4 from proof inj (previous step)

n→i(m) = n→i(n0) + 1, ∀ m. n→i(m) = n→i(n0) → m = n0 ⊦ m = n0 + 1

User applied rule all left

And obtained the following new goals

Proof step 5:
n→i(m) = n→i(n0) + 1, n→i(m -1) = n→i(n0) → m -1 = n0, ∀ m. n→i(m) = n→i(n0) → m = n0 ⊦ m = n0 + 1

Rule argument

pos: left 2 terms: m -1 abort: F discard: F computed: F

Additional information

Used terms: m -1