Metamath Proof Explorer
Description: A lemma for eliminating inequality, in inference form. (Contributed by Giovanni Mascellani, 31-May-2019)
|
|
Ref |
Expression |
|
Hypotheses |
sbceq1ddi.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
|
sbceq1ddi.2 |
⊢ ( 𝜓 → 𝜃 ) |
|
|
sbceq1ddi.3 |
⊢ ( [ 𝐴 / 𝑥 ] 𝜒 ↔ 𝜃 ) |
|
|
sbceq1ddi.4 |
⊢ ( [ 𝐵 / 𝑥 ] 𝜒 ↔ 𝜂 ) |
|
Assertion |
sbceq1ddi |
⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜂 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
sbceq1ddi.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
| 2 |
|
sbceq1ddi.2 |
⊢ ( 𝜓 → 𝜃 ) |
| 3 |
|
sbceq1ddi.3 |
⊢ ( [ 𝐴 / 𝑥 ] 𝜒 ↔ 𝜃 ) |
| 4 |
|
sbceq1ddi.4 |
⊢ ( [ 𝐵 / 𝑥 ] 𝜒 ↔ 𝜂 ) |
| 5 |
1
|
adantr |
⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝐴 = 𝐵 ) |
| 6 |
2 3
|
sylibr |
⊢ ( 𝜓 → [ 𝐴 / 𝑥 ] 𝜒 ) |
| 7 |
6
|
adantl |
⊢ ( ( 𝜑 ∧ 𝜓 ) → [ 𝐴 / 𝑥 ] 𝜒 ) |
| 8 |
5 7
|
sbceq1dd |
⊢ ( ( 𝜑 ∧ 𝜓 ) → [ 𝐵 / 𝑥 ] 𝜒 ) |
| 9 |
8 4
|
sylib |
⊢ ( ( 𝜑 ∧ 𝜓 ) → 𝜂 ) |