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Proof step 2 from proof postfix-trans (previous step)

x ⊒ x0, x0 ⊒ y ⊦ x ⊒ y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ x ⊒ y ↔ (∃ z. z + x = y)
Proof step 4:
x ⊒ x0, x0 ⊒ y ⊦ ∃ z. z + x = y

Rule argument

spec: inst: name: postfix seq: ⊦ x ⊒ y ↔ (∃ z. z + x = y) subst: [y, x] → [y, x] flags: rewrite, current

Additional information

Used terms: y, x