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Proof step 5 from proof ex (previous step)

∀ a. ∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ x0 = y + z ⊦ ∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ a0 ' + x0 = y + z

User applied rule all left

And obtained the following new goals

Proof step 6:
∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ x0 = y + z, ∀ a. ∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ x0 = y + z ⊦ ∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ a0 ' + x0 = y + z

Rule argument

pos: left 1 terms: a abort: F discard: F computed: F

Additional information

Used terms: a